[tex]\begin{array}{|c|ll} \cline{1-1} \textit{\textit{\LARGE a}\% of \textit{\LARGE b}}\\ \cline{1-1} \\ \left( \cfrac{\textit{\LARGE a}}{100} \right)\cdot \textit{\LARGE b} \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{3.8\% of 25}}{\left( \cfrac{3.8}{100} \right)25}\implies \text{\LARGE 0.95}[/tex]
Answer:
To find 3.8% of 25, we can convert the percentage to a decimal and then multiply it by 25:
3.8% = 0.038 (converting percentage to decimal)
0.038 x 25 = 0.95
Therefore, 3.8% of 25 is 0.95.
degree = 7, zeroes of multiplicity 2 at x=3 and x=-4, a zero of mutiplicity at x=-1 and y-intercept =(0,4)
The equation of the polynomial is f(x) = (1/36)(x - 3)2(x + 4)2(x + 1)3.
The equation of the polynomial can be found using the given information about the degree, zeroes, and y-intercept. The general form of a polynomial is:
f(x) = a(x - x1)n1(x - x2)n2...(x - xk)nk
Where x1, x2, ..., xk are the zeroes of the polynomial, n1, n2, ..., nk are the multiplicities of the zeroes, and a is the leading coefficient.
Using the given information, we can plug in the values for the zeroes and their multiplicities:
f(x) = a(x - 3)2(x + 4)2(x + 1)3
The degree of the polynomial is 7, which is the sum of the multiplicities of the zeroes. So, the equation is correct in terms of the degree.
Now, we need to find the value of a using the y-intercept. The y-intercept is the point where the graph of the polynomial crosses the y-axis, which means that x = 0. So, we can plug in x = 0 and the given y-intercept (0,4) into the equation and solve for a:
4 = a(0 - 3)2(0 + 4)2(0 + 1)3
4 = a(9)(16)(1)
4 = 144a
a = 4/144
a = 1/36
Now, we can plug in the value of a back into the equation to get the final answer:
f(x) = (1/36)(x - 3)2(x + 4)2(x + 1)3
Therefore, the equation of the polynomial is f(x) = (1/36)(x - 3)2(x + 4)2(x + 1)3.
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Solve for y. 3y^(2)=7y+6 If there is more than one solution, separate them wit If there is no solution, click on "No solution."
The solutions to the equation are y = -2/3 and y = 3.
To solve for y, we need to rearrange the equation and use the quadratic formula. Here are the steps:
1. Subtract 7y and 6 from both sides of the equation to get: 3y^(2) - 7y - 6 = 0
2. Use the quadratic formula to solve for y: y = (-b ± √(b^(2) - 4ac))/(2a), where a = 3, b = -7, and c = -6
3. Plug in the values for a, b, and c into the formula and simplify: y = (-(-7) ± √((-7)^(2) - 4(3)(-6)))/(2(3))
4. Simplify further: y = (7 ± √(49 + 72))/6
5. Simplify the square root: y = (7 ± √121)/6
6. Simplify further: y = (7 ± 11)/6
7. Solve for the two possible values of y: y = (7 + 11)/6 = 18/6 = 3, and y = (7 - 11)/6 = -4/6 = -2/3
Therefore, the two solutions for y are 3 and -2/3.
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? QUESTION Use the rational zeros theorem to list all possible rational zeros of the following. h(x)=-2x^(4)+7x^(3)-8x^(2)+9x+3 Be sure that no value in your list appears more than once.
The possible rational zeros of h(x) are 1, -1, 3, -3, 1/2, -1/2, 3/2, and -3/2.
Rational Zeros TheoremAccording to the Rational Zeros Theorem, if a polynomial has any rational zeros, they must be in the form of p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.
For the given polynomial h(x)=-2x^(4)+7x^(3)-8x^(2)+9x+3, the constant term is 3 and the leading coefficient is -2.
The factors of 3 are 1 and 3, and the factors of -2 are 1, 2, -1, and -2.
Therefore, the possible rational zeros of h(x) are: p/q = ±1/1, ±3/1, ±1/2, ±3/2
Simplifying these values, we get the following list of possible rational zeros:
±1, ±3, ±1/2, ±3/2
So, the possible rational zeros of h(x) are 1, -1, 3, -3, 1/2, -1/2, 3/2, and -3/2.
Be sure to check each of these values by plugging them back into the original polynomial to see if they are actually zeros.
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Image attached - thanks for helping!
Answer:
Ans = (1.50 x 29 movies) + 16
Step-by-step explanation:
y represents total costs
x represents how many movie he rents
$1.50 is the rent of each movie
$16 is the fixed fee
thus
y = 1.50 x + 16
Ans = (1.50 x 29 movies) + 16
A chemist has 42 grams of aluminum. There are 2.70 grams/milliliter for aluminum. How many milliliters of aluminum does the chemist have? Set this up either as a proportion or unit analysis.
a. the chemist has 133 milliliters
b. the chemist has 155 milliliters
c. the chemist has 58 milliliters
d. the chemist has 16 milliliters
The 42 grams of aluminum implies that the chemist has 0.06429ml of aluminum
What is density ?The density of material shows the denseness of that material in a specific given area. A material’s density is defined as its mass per unit volume. Density is essentially a measurement of how tightly matter is packed together. It is a unique physical property of a particular object. The principle of density was discovered by the Greek scientist Archimedes. It is easy to calculate density if you know the formula and understand the related units The symbol ρ represents density or it can also be represented by the letter D.
How to determine the amount of aluminum?
The mass is given as: Mass = 42 grams
The density of aluminum is: Density = 2.7 g/cm³
So, we have: Volume = Density/Mass
This gives, Volume = 2.7/42
Evaluate : Volume = 0.06429cm³
Convert to ml
Volume = 0.06429ml
Hence, the chemist has 0.06429ml of aluminum
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How much oil should Kim use? Complete the table
Kim should use 8/3 (or 2.67) ounces of oil in the recipe.
what is a linear equation?
A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a variable raised to the first power (i.e., the exponent of the variable is 1).
Based on the table, we can see that the recipe calls for a total of 24 ounces of dressing, with a ratio of 3 parts oil to 1 part vinegar.
To calculate how much oil Kim should use, we can set up a proportion:
3 parts oil : 1 part vinegar = x ounces of oil : 8 ounces of vinegar
Cross-multiplying, we get:
3x = 8
Dividing both sides by 3, we get:
x = 8/3
To complete the table, we can fill in the remaining values based on the given ratio:
Total Ounces of Dressing Ounces of Oil Ounces of Vinegar
12 9 3
16 12 4
20 15 5
24 18 6
Therefore, Kim should use 8/3 (or 2.67) ounces of oil in the recipe.
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The table below shows the ages of 50 people on a bus.
Age (years) Frequency
0 < h ≤ 20 20
20 < h ≤ 40 10
40 < h ≤ 60 10
60 < h ≤ 80 10
Calculate an estimate for the mean age.
An estimate for the mean age of the 50 people on the bus is 34 years.
What is Statistics?Statistics is the discipline that concerns the collection, organization, analysis, interpretation, and presentation of data.
Let us calculate the sum of the products of the midpoints and their respective frequencies, and divide by the total frequency to obtain the estimated mean age.
The midpoint and frequency data for each age group can be represented as follows:
Age group Midpoint Frequency
0 < h ≤ 20 10 20
20 < h ≤ 40 30 10
40 < h ≤ 60 50 10
60 < h ≤ 80 70 10
To calculate the estimated mean age, we can use the formula:
Estimated mean age = (Σ(midpoint × frequency)) / total frequency
Estimated mean age = ((10 × 20) + (30 × 10) + (50 × 10) + (70 × 10)) / (20 + 10 + 10 + 10)
Estimated mean age = (200 + 300 + 500 + 700) / 50
= 1700 / 50
= 34
Therefore, an estimate for the mean age of the 50 people on the bus is 34 years.
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The Johnsons framed a family picture to hang on the wall. The perimeter of the frame is 42 inches.
By answering the above question, we may state that We are unable to solve this equation for both l and w since it has only one solution and two unknowns.
what is function?Mathematicians research numbers, their variants, equations, forms, and related structures, as well as possible locations for these things. The relationship between a group of inputs, each of which has a corresponding output, is referred to as a function. Every input contributes to a single, distinct output in a connection between inputs and outputs known as a function. A domain, codomain, or scope is assigned to each function. Often, functions are denoted with the letter f. (x). The key is an x. There are four main categories of accessible functions: on functions, one-to-one capabilities, so many capabilities, in capabilities, and on functions.
We are aware of the frame's 42-inch circumference but are unaware of the picture's or frame's shape, as well as if the frame's breadth fluctuates or is constant.
With the assumption that the frame is rectangular and has a constant width, we might create the equation shown below:
2l + 2w + 4x = 42
where L stands for the picture's length, W for its width, and X for the frame's width (assuming there are four sides with the same width).
We are unable to solve this equation for both l and w since it has only one solution and two unknowns.
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hen writing in row-reduced echelon form -15x-10y-20z=-6 9x+12y+30z=15 3x+2y+10z=6
The solution to the given system of equations is:
x = -4/7
y = -9/14
z = 18/7
To write the given system of equations in row-reduced echelon form, we will first need to convert the equations to an augmented matrix, and then use row operations to reduce the matrix to its row-reduced echelon form.
The augmented matrix for the given system of equations is:
| -15 -10 -20 | -6 |
| 9 12 30 | 15 |
| 3 2 10 | 6 |
We will now use row operations to reduce the matrix to its row-reduced echelon form:
Divide the first row by -15 to get a leading 1 in the first row:
| 1 2/3 4/3 | 2/5 |
| 9 12 30 | 15 |
| 3 2 10 | 6 |
Subtract 9 times the first row from the second row, and 3 times the first row from the third row:
| 1 2/3 4/3 | 2/5 |
| 0 2 6 | 3 |
| 0 -4/3 -2/3 | 18/5 |
Divide the second row by 2 to get a leading 1 in the second row:
| 1 2/3 4/3 | 2/5 |
| 0 1 3 | 3/2 |
| 0 -4/3 -2/3 | 18/5 |
Add 4/3 times the second row to the third row:
| 1 2/3 4/3 | 2/5 |
| 0 1 3 | 3/2 |
| 0 0 14/3 | 54/5 |
Divide the third row by 14/3 to get a leading 1 in the third row:
| 1 2/3 4/3 | 2/5 |
| 0 1 3 | 3/2 |
| 0 0 1 | 18/7 |
Subtract 3 times the third row from the second row, and 4/3 times the third row from the first row:
| 1 2/3 0 | -16/21 |
| 0 1 0 | -9/14 |
| 0 0 1 | 18/7 |
Subtract 2/3 times the second row from the first row:
| 1 0 0 | -4/7 |
| 0 1 0 | -9/14 |
| 0 0 1 | 18/7 |
The row-reduced echelon form of the given system of equations is:
| 1 0 0 | -4/7 |
| 0 1 0 | -9/14 |
| 0 0 1 | 18/7 |
So, the solution to the given system of equations is:
x = -4/7
y = -9/14
z = 18/7
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Your current test average in Science is an 81. You've taken 5 tests. You only have one test remaining. What grade would you need on the
next test to raise your average to an 85?
Answer:
You must get a 89%
Step-by-step explanation:
81+89=170
170/2= 85%
Joseph places $5,500 in a savings account for 30 months. He earns $893.75 in interest. What is the annual interest rate?
The annual interest rate is 6.5%
How to calculate the annual interest rate?The first step is to write out the parameters given in the question
Joseph places $5500 in a savings account for 30 months
He rans $893.75 in interest
The annual interest rate can be calculated by multiplying the amount in the savings by the number of months which is 30
= 5500 × 30y
= 165,000y
= 165,000y/12
= 13,750y
Equate 13,750y with the amount of interest
13,750y= 898.75
y= 898.75/ 13,750
y= 0.065 × 100
= 6.5%
Hence the annual interest rate is 6.5%
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Which number is a perfect square
2
10
50
100
Answer:
100
Step-by-step explanation:
hope this helps! :))
Answer: 100
Step-by-step explanation:
A perfect square can be found when you multiply the same integer twice. 100 is a perfect square because 10 * 10 = 100. You multiply 10 twice in the equation.
21. Suppose that a given population can be divided into two parts: those who have given disease and can infect Others, and those who do not have it but are susceptible. Let x be the proportion of susceptible individuals and y the proportion of infectious individuals; then x + y = 1. Assume that the disease spreads by contact between sick and well members of the population and that the rate of spread dy/dt is proportional to the number of such contacts. Further; assume that members of both groups move about freely among each other; so the number of contacts is proportional to the product ofx and y. Since =1 -Y, we obtain the initial value problem dy = ay( 1 ~y), y(0) = Jo; (22) dt where is positive proportionality factor, and Yo is the initial proportion of infectious individuals. A. Find the equilibrium points for the differential equation (22) and determine whether each is asymptotically stable, semistable. Or unstable. B. Solve the initial value problem 22 and verify that the conelusions You reached in part a are correet: Show that y(t) as 5 C, which means that ultimately the disease spreads through the entire population
The Initial value problem is defined
as [tex] \frac{dy}{dt} =\alpha y(1 -y), y(0) = y_0[/tex]
a) Equilibrium points or critical values are y = 0 and y = 1. Also, y = 0, is unstable and y = 1, is asymptomatic stable.
b) The solution of above initial value problem is y = 1 , which means at the end the disease will spread through the entire population.
We have a population data which can be divided into two parts. Let consider x be the proportion of susceptible individuals and y the proportion of infectious individuals; then x + y = 1. Now, Initial value problem ( that a differential equation) is, [tex] \frac{dy}{dt} = \alpha y(1 -y), y(0) = y_0[/tex]
a) we have to determine equilibrium points and nature of asymptote for above equation. To determine the equilibrium solution of equation we must put, dy/dt = 0, for all t values. At equilibrium, dy/dt = 0
=> αy(1 - y ) = 0
=> y( 1 - y) = 0
=> either y = 0 or 1 - y = 0
=> y = 0 or y = 1
so, y = 0 is unstable and y = 1 , asymptomatic stable.
b) Now, we have to solve initial value problem, [tex]\frac{dy}{y(1 - y)} = \alpha dt [/tex]
Using partial fraction decomposition,
[tex] \frac{1}{y(1 - y) }= \frac{1}{y} - \frac{1}{1 - y}[/tex]
integrating both sides,
[tex]\int {( \frac{1}{y} - \frac{1}{(1 - y)})dy } = \int{ \alpha dt }[/tex]
[tex]ln (y) - ln (1 - y) = \alpha t + c[/tex]
[tex] ln( \frac{ y}{1- y}) \: = \alpha t + c [/tex]
[tex] \frac{y}{1 - y}= e^{\alpha t} c_1 [/tex]
using initial condition, [tex]y(0) = y_0 [/tex]
[tex] \frac{y_0}{1 - y_0}= 1 ×c_1 [/tex]
[tex]c_1 = \frac{ y_0}{1 - y_0}[/tex]
[tex]so, \frac{y}{1 - y} = \frac{ y_0}{1 - y_0}e^{\alpha t} [/tex]
cross multiplication
[tex]y(1 - y_0) = ((1 - y) y_0 )e^{\alpha t} [/tex]
[tex]y - yy_0 = y_0e^{\alpha t} - yy_0 e^{\alpha t} [/tex]
[tex]y = \frac{ y_0}{y_0 + ( 1 - y_0) e^{-\alpha t} }[/tex]
as [tex]t→ ∞ , e^{- \alpha t } → 0[/tex]
[tex]y = \frac{ y_0}{y_0 }= 1 [/tex]
So, y = 1, means that ultimately the disease spreads through the entire population.
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What is 5x + 4y = -16 in slope-intercept form?
Using Cramer's Rule In Exercises 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, and 22, use Cramer's Rule to solve (if possible) the system of linear equations. 13x - 6у = 17 26x -12y = 8
Using Cramer's Rule, we cannot solve this system of linear equations because the determinant of the coefficient matrix is 0. However, we can conclude that the system has infinitely many solutions.
Using Cramer's Rule, we can solve the system of linear equations 13x - 6у = 17 and 26x -12y = 8 by following these steps:
Step 1: Find the determinant of the coefficient matrix, which is:
| 13 -6 |
| 26 -12 |
= (13)(-12) - (-6)(26)
= -156 - (-156)
= 0
Step 2: Since the determinant of the coefficient matrix is 0, we cannot use Cramer's Rule to solve this system of equations.
Step 3: We can check if the system has no solution or infinitely many solutions by comparing the ratios of the coefficients of x and y in the two equations.
The ratio of the coefficients of x in the two equations is 13/26 = 1/2, and the ratio of the coefficients of y in the two equations is -6/-12 = 1/2. Since these ratios are equal, the system has infinitely many solutions.
Using Cramer's Rule, we cannot solve this system of linear equations because the determinant of the coefficient matrix is 0. However, we can conclude that the system has infinitely many solutions.
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Evaluate det(A) by a cofactor expansion along a row or column of your choice. 5 0 0 1 0 3 3 3 -1 0 A=1 5 4 2 3 4 4 2 2 3 2 54 2 3 A = i
To evaluate det(A) by a cofactor expansion along a row or column of our choice, we will use the following formula:
det(A) = a11C11 + a12C12 + a13C13 + ... + a1nC1n
where aij is the element in the ith row and jth column of matrix A, and Cij is the cofactor of that element.
For simplicity, we will choose to expand along the first row of matrix A.
det(A) = 5C11 + 0C12 + 0C13 + 1C14 + 0C15 + 3C16 + 3C17 + 3C18 + (-1)C19 + 0C110
Now, we will find the cofactors of each element in the first row.
C11 = (-1)1+1det(M11) = det(M11) = (5)(4)(3)(54) - (3)(2)(4)(3) = 3240 - 72 = 3168
C12 = (-1)1+2det(M12) = -det(M12) = -(0)
C13 = (-1)1+3det(M13) = det(M13) = (0)
C14 = (-1)1+4det(M14) = det(M14) = (3)(4)(3)(54) - (2)(2)(4)(3) = 1944 - 48 = 1896
C15 = (-1)1+5det(M15) = -det(M15) = -(0)
C16 = (-1)1+6det(M16) = det(M16) = (2)(4)(3)(54) - (2)(2)(4)(3) = 1296 - 48 = 1248
C17 = (-1)1+7det(M17) = -det(M17) = -(2)(3)(3)(54) - (2)(2)(4)(3) = -972 - 48 = -1020
C18 = (-1)1+8det(M18) = det(M18) = (1)(3)(3)(54) - (2)(2)(4)(3) = 486 - 48 = 438
C19 = (-1)1+9det(M19) = -det(M19) = -(1)(4)(3)(54) - (2)(2)(4)(3) = -648 - 48 = -696
C110 = (-1)1+10det(M110) = det(M110) = (0)
Now, we will substitute the cofactors back into the formula and simplify.
det(A) = 5(3168) + 0(0) + 0(0) + 1(1896) + 0(0) + 3(1248) + 3(-1020) + 3(438) + (-1)(-696) + 0(0)
det(A) = 15840 + 1896 + 3744 - 3060 + 1314 + 696
det(A) = 19430
Therefore, det(A) = 19430.
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Pick three times (independently) a point at random from the interval (0,1). a. Let X be the number of picked points that is smaller than 1/4. Determine the distribution of X. b. Let Y be the middle one of the three points. Determine the cdf of Y. (It is a function with domain R!) c. Determine the pdf of Y.
Y(y) = dF_Y(y)/dy = -y^2
a. Let X be the number of picked points that is smaller than 1/4. Determine the distribution of X.Random variables and distributions Let A be the set of all possible 3-tuples (a1, a2, a3) picked from (0, 1).The number of the picked points that are smaller than 1/4 is random with X: A → {0, 1, 2, 3}X((a1, a2, a3)) = |{i ∈ {1, 2, 3} : ai < 1/4}|This is, X is the number of i's such that ai < 1/4. We would like to compute the distribution of X.Let us count the number of 3-tuples (a1, a2, a3) for which X = 0, 1, 2, or 3. These counts are as follows:X = 0: There is only one tuple for which all ai ≥ 1/4.X = 1: Each of the ai can be either ≥ 1/4 or < 1/4, and there are three ways to choose which one is less than 1/4. Therefore, there are 2^3 · 3 = 24 3-tuples (a1, a2, a3) such that exactly one ai < 1/4.X = 2: Either all ai < 1/4, or two ai's < 1/4 and the other one ≥ 1/4. If all ai < 1/4, then there is one tuple for this. If two ai's < 1/4 and the other one ≥ 1/4, then there are 3 ways to choose which ai is ≥ 1/4, and for each such choice there are 3 · 2 = 6 ways to pick the other two ai's. Thus, there are 1 + 3 · 6 = 19 tuples for which X = 2.X = 3: All ai's < 1/4. There is only one such tuple.Therefore, the probability mass function of X is as follows:P(X = 0) = 1/8P(X = 1) = 3/8P(X = 2) = 19/32P(X = 3) = 1/32b. Let Y be the middle one of the three points. Determine the cdf of Y. (It is a function with domain R!)The range of Y is the interval (0, 1). We can assume that a1 ≤ a2 ≤ a3, since any 3-tuple can be sorted in this way. If Y < y, then the largest of the three points must be less than y. Therefore, we need to find the probability that the largest point is less than y, given that the three points are picked independently at random from (0, 1) and are ordered as a1 ≤ a2 ≤ a3.Let A be the set of all possible 3-tuples (a1, a2, a3) picked from (0, 1), where a1 ≤ a2 ≤ a3. Let B(y) be the set of all 3-tuples (a1, a2, a3) ∈ A such that a3 < y. Then P(Y < y) = P(B(y)).To compute P(B(y)), let us count the number of 3-tuples (a1, a2, a3) ∈ A such that a3 < y. Fix a1 and a2. Then a3 < y if and only if a3 ∈ (0, y), so there are exactly y(1 - y) choices for a3. Therefore,|B(y)| = ∫∫A y(1 - y) da1 da2 = ∫0^1 ∫a1^1 y(1 - y) da2 da1 = ∫0^1 y(1 - y) (1 - a1) da1 = 1/6 - y^3/3Thus,P(Y < y) = P(B(y)) = |B(y)|/|A| = [1/6 - y^3/3]/1 = 1/6 - y^3/3This is the cdf of Y for y ∈ (0, 1). We can extend this function to all of R by setting F_Y(y) = 0 if y ≤ 0 and F_Y(y) = 1 if y ≥ 1.c. Determine the pdf of Y.The cdf of Y isF_Y(y) = 1/6 - y^3/3for y ∈ (0, 1). Therefore, the pdf of Y isf_Y(y) = dF_Y(y)/dy = -y^2for y ∈ (0, 1). Thus, the density is constant, and Y has a uniform distribution on (0, 1).
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Work out the area of trapezium H.
If your answer is a decimal, give it to 1 d.p.
URGENTLY NEED HELP
Answer:
Step-by-step explanation:
I don't know if this is right but i'm just trying to help you. So I did 27 times 9 because the blue part is 9cm and the pink part H is 27cm, I got 243. Then the whole thing is 243cm but you see how it says in blue area=30cm well that means that that small peace if 30cm out of the whole 243cm, so what I did was i subtracted 243cm by 30cm and I got a answer of 213 cm. If that was what you had needed help on. I hope i helped you even a little bit, please let me know if i did help you. And also um i don't know how to friend someone so if you can tell me how to friend someone i was thinking about friending you. And ill also just you some points if you want, say mabey like 10 or 20 points. Thanks and hoped that helped and good luck buddy. :)
Find the measurement of AC to nearest inch.
Answer:
AC is about 33 inches.
Step-by-step explanation:
Set your calculator to degree measure.
[tex] \tan(35) = \frac{x}{47} [/tex]
[tex]x = 47 \tan(35) = 32.91[/tex]
List the members of the following sets: (2 marks)
y|y=(x/x+1), x ϵ Z+, x<12) (Hint: 0 is considered a positive integer)
{x | x is the square of a whole number and x < 100} (Hint: Whole numbers include all positive integers along with zero)
The members of the second set are 0, 1, 4, 9, 16, 25, 36, 49, 64, and 81.
The members of the first set, y|y=(x/x+1), x ϵ Z+, x<12, are the values of y that satisfy the given equation and conditions. The members of the second set, {x | x is the square of a whole number and x < 100}, are the values of x that satisfy the given conditions.
For the first set, we need to find the values of y that satisfy the equation y=(x/x+1) for positive integers x less than 12. The values of x that satisfy this condition are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11. We can plug these values into the equation to find the corresponding values of y:
y = (1/1+1) = 0.5
y = (2/2+1) = 0.6667
y = (3/3+1) = 0.75
y = (4/4+1) = 0.8
y = (5/5+1) = 0.8333
y = (6/6+1) = 0.8571
y = (7/7+1) = 0.875
y = (8/8+1) = 0.8889
y = (9/9+1) = 0.9
y = (10/10+1) = 0.9091
y = (11/11+1) = 0.9167
Therefore, the members of the first set are 0.5, 0.6667, 0.75, 0.8, 0.8333, 0.8571, 0.875, 0.8889, 0.9, 0.9091, and 0.9167.
For the second set, we need to find the values of x that are the square of a whole number and less than 100. The whole numbers that satisfy this condition are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. We can square these values to find the corresponding values of x:
x = 0^2 = 0
x = 1^2 = 1
x = 2^2 = 4
x = 3^2 = 9
x = 4^2 = 16
x = 5^2 = 25
x = 6^2 = 36
x = 7^2 = 49
x = 8^2 = 64
x = 9^2 = 81
Therefore, the members of the second set are 0, 1, 4, 9, 16, 25, 36, 49, 64, and 81.
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The length of a rectangle is 2 inches more than its width. Write an equation relating the length l of the rectangle to its width w
Answer: If the length of a rectangle is 2 inches more than its width, we can write:
length = width + 2
We can also rearrange this equation to solve for the width in terms of the length:
width = length - 2
Both of these equations relate the length l of the rectangle to its width w.
Step-by-step explanation:
According to a survey of workers,
workers walk or bike to work?
7/25
of them walk to work,
2/25
bike,
4/25
carpool, and
12/25
drive alone.
The percentage of workers who likes to walk or bike to work is P = 36 %
What is Percentage?A percentage is a number or ratio expressed as a fraction of 100. It is often denoted using the percent sign, %
The difference between an exact value and an approximation to it is the approximation error in a data value. Either an absolute error or a relative error might be used to describe this error.
Percentage change is the difference between the measured value and the true value , as a percentage of the true value
Percentage change =( (| Measured Value - True Value |) / True Value ) x 100
Given data ,
Let the percentage of workers who likes to walk or bike to work be P
The percentage of workers who likes to walk to work = 7/25 = 28/100
The percentage of workers who likes to bike to work = 2/25 = 8/100
The percentage of workers who likes to carpool to work = 4/25 = 16/100
The percentage of workers who likes to drive to work = 12/25 = 48/100
So , the percentage of workers who likes to walk or bike is given by P
where P = 28/100 + 8/100
On simplifying , we get
The percentage of workers who likes to walk or bike to work P = 36/100
The percentage of workers who likes to walk or bike to work P = 36 %
Hence , the percentage of workers is 36 %
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Mr. Smith is putting a brick border around his irregular shaped yard. Before installing the
border, he must cut the bricks to fit the angles of the garden. Use the given measures to answer
the question.
∡ = (7 − 8)°
∡ = (4 + 46)°
∡ = (5)°
∡ = (6 + 12)°
∡ = ( + 7)°
∡ = (5 − 9)°
What are the angle measures of each vertex of the yard? Show your work
All the angle measures are as follows: m∠A = 160°, m∠B = 142°, m∠C = 120°, m∠D = 156°, m∠E = 31°, m∠F = 111°
We know that a hexagon is a six-sided polygon and also a convex hexagon which basically means that the hexagon's vertices are pointed outwards.
And with no interior angle has an angle measuring of more than 180°.
We know that the number of sides in the polygon = 6.
And the total sum of the irregular angle = (4 - 2) x 180° = 720°.
7x - 8 + 4x + 46 + 5x + 6x + 12 + x + 7 + 5x - 9 = 720°.
28x = 720 - 48
x = 24
Substituting all the values of x = 24 to all the interior angle measures,
Now, all the measures are:
m∠A = 7x - 8 = 160°
m∠B = 4x + 46 = 142°
m∠C = 5x = 120°
m∠D = 6x + 12 = 156°
m∠E = x + 7 = 31°
m∠F = 5x - 9 = 111°
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Name Identities Idempotent laws A u A = A AnA = A Associative laws (AU B) U C = Au (B U C) (AnB) nc =An(Bnc) Commutative laws Au B = B U A AnB = BnA Distributive laws Au (B n C) = (Au B) n (Au C) An(B u C) (AnB) u (An C) Identity laws Au 0 = A AnU = A Domination laws Ano =0 Au U = U Double Complement law A = A AnA = U = 0 Au A = U 0 = U Complement laws De Morgans laws AuB = AnB AnB = Au B Absorption laws Au (AnB) = A An(A U B) = A
The Idempotent laws, Associative laws, Commutative laws, Distributive laws, Identity laws, Domination laws, Double Complement law, Complement laws, and De Morgan's laws are all types of identities that are used in set theory and Boolean algebra. These identities are used to simplify expressions and to prove the equivalence of two expressions.
The Idempotent laws state that the union or intersection of a set with itself is equal to the set itself:
A u A = A
AnA = A
The Associative laws state that the order in which sets are grouped does not matter when taking the union or intersection:
(AU B) U C = Au (B U C)
(AnB) nc =An(Bnc)
The Commutative laws state that the order in which sets are listed does not matter when taking the union or intersection:
Au B = B U A
AnB = BnA
The Distributive laws state that the union or intersection of a set with the intersection or union of two other sets is equal to the intersection or union of the set with each of the other sets:
Au (B n C) = (Au B) n (Au C)
An(B u C) (AnB) u (An C)
The Identity laws state that the union of a set with the empty set is equal to the set itself, and the intersection of a set with the universal set is equal to the set itself:
Au 0 = A
AnU = A
The Domination laws state that the intersection of a set with the empty set is equal to the empty set, and the union of a set with the universal set is equal to the universal set:
Ano =0
Au U = U
The Double Complement law states that the complement of the complement of a set is equal to the set itself:
A = A
AnA = U = 0
Au A = U 0 = U
The Complement laws state that the union of a set with its complement is equal to the universal set, and the intersection of a set with its complement is equal to the empty set:
AuB = AnB
AnB = Au B
De Morgan's laws state that the complement of the union of two sets is equal to the intersection of the complements of the sets, and the complement of the intersection of two sets is equal to the union of the complements of the sets:
AuB = AnB
AnB = Au B
The Absorption laws state that the union of a set with the intersection of the set with another set is equal to the set itself, and the intersection of a set with the union of the set with another set is equal to the set itself:
Au (AnB) = A
An(A U B) = A
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You got home on day two in 75% of the time it took you on day one. On day three it took you 20% longer than on day two. On day four it took you two minutes less than on day three. On day five you walked 50% faster than on day four and it took you 15 minutes to get home. How long did it take you to get home on day one?
So the answer is: it took 35.56 minutes to get home on day one.
To find out how long it took to get home on day one, we need to work backwards from day five. Here's how:
On day five, it took 15 minutes to get home. This was 50% faster than on day four, which means that on day four it took 15 minutes / 0.5 = 30 minutes.
On day four, it took 30 minutes, and this was 2 minutes less than on day three. So on day three it took 30 minutes + 2 minutes = 32 minutes.
On day three, it took 32 minutes, and this was 20% longer than on day two. So on day two it took 32 minutes / 1.2 = 26.67 minutes.
On day two, it took 26.67 minutes, and this was 75% of the time it took on day one. So on day one it took 26.67 minutes / 0.75 = 35.56 minutes.
So the answer is: it took 35.56 minutes to get home on day one.
Here's the answer formatted with HTML:
To find out how long it took to get home on day one, we need to work backwards from day five. Here's how:
So the answer is: it took 35.56 minutes to get home on day one.
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Find the point in which the line through the origin perpendicular to the plane 2x - y - z = 4 meets the plane 3x - 5y + 2z = 6. - . - -
The point in which the line through the origin perpendicular to the plane 2x - y - z = 4 meets the plane 3x - 5y + 2z = 6 is (4/3, -2/3, -2/3).
To find the point in which the line through the origin perpendicular to the plane 2x - y - z = 4 meets the plane 3x - 5y + 2z = 6, we can use the following steps:
Find the normal vector of the plane 2x - y - z = 4. This is given by the coefficients of the x, y, and z terms, which are 2, -1, and -1, respectively. So the normal vector is (2, -1, -1).
Since the line through the origin is perpendicular to the plane, it must be parallel to the normal vector of the plane. Therefore, the direction vector of the line is (2, -1, -1).
The equation of the line through the origin with direction vector (2, -1, -1) is given by x = 2t, y = -t, and z = -t, where t is a parameter.
Substitute the equations of the line into the equation of the plane 3x - 5y + 2z = 6 to find the value of t:
3(2t) - 5(-t) + 2(-t) = 6
6t + 5t - 2t = 6
9t = 6
t = 2/3
Substitute the value of t back into the equations of the line to find the point of intersection:
x = 2(2/3) = 4/3
y = -(2/3) = -2/3
z = -(2/3) = -2/3
Therefore, the point in which the line through the origin perpendicular to the plane 2x - y - z = 4 meets the plane 3x - 5y + 2z = 6 is (4/3, -2/3, -2/3).
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Determine which method would be best to solve this system. Then solve. 3x-2y=4 2x-2y=-8 a. elimination using subtraction; (12,10) b. elimination using addition: (2,6) c. elimination using subtraction;
The best method to solve this system is elimination using subtraction and the solution is (12, 16).
First, we need to get one of the variables to have the same coefficient in both equations so we can eliminate it. In this case, the y variable already has the same coefficient of -2 in both equations.
Next, we subtract the second equation from the first equation to eliminate the y variable:
3x - 2y = 4
-(2x - 2y = -8)
This gives us:
x = 12
Now, we can plug this value of x back into one of the original equations to solve for y:
3(12) - 2y = 4
36 - 2y = 4
-2y = -32
y = 16
So the solution to this system is (12, 16).
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To earn their next scouting badge, Craig and Quinn are working on a STEM project. They're building a tower with toothpicks and miniature marshmallows. For every stories they build their tower, they use miniature marshmallows. Complete the table
According to the information, the values that complete the table are: 4, 6, and 150.
How to calculate the missing numbers in the table?To calculate the missing numbers in the table we must analyze the existing data. Once we have analyzed this data we can establish that it is a constant relationship, so we can find the missing numbers by means of a rule of three as shown below:
30 = 2
60 =?
60 * 2/30 = 4
30 = 2
90 =?
90 * 2/30 = 6
2 = 30
10 =?
10 * 30/2 = 150
According to the above, the missing values in the box from left to right are: 4, 6, and 150.
Note: This question is incomplete. Here is the complete information:
Attached image
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help me complete this and i will mark brainlirst
Answer:
Step-by-step explanation:
The method to find the midpoint of two points is (a + b)/2 a and b both being x+x and y+y for the differentiating coordinates respectively.
here, it would be x = (-2 + 4)/2 and y = (5-5)/2
giving you the coordinate [ 1 , 0 ]
Evaluating Trigonometric Functions In Exercises 15-30, find the exact values of the sine, cosine, and tangent of the angle.
15. 105 deg = 60 deg + 45 deg
17. 195 deg = 225 deg - 30 deg
19. (11pi)/12 = (3pi)/4 + pi/6
12 21. - pi/12 = pi/6 - pi/4
23. 75°
25.-285°
27. (13pi)/12
222 29. - (7pi)/12
16. 165 deg = 135 deg + 30 deg
18. 255 deg = 300 deg - 45 deg
20. (17pi)/12 = (7pi)/6 + pi/4
22. - (19Y)/12 = (2pi)/3 - (9pi)/4
24. 15°
26.-165°
28. (5pi)/12
30. - (13pi)/12
For 15:
Sine: √2/2
Cosine: √2/2
Tangent: 1
For 17:
Sine: -1/2
Cosine: √3/2
Tangent: -1/√3
For 19:
Sine: (1+√3)/2
Cosine: (1-√3)/2
Tangent: √3
For 21:
Sine: -1/2
Cosine: -√3/2
Tangent: 1/√3
For 23:
Sine: √3/2
Cosine: 1/2
Tangent: √3
For 25:
Sine: -√2/2
Cosine: -√2/2
Tangent: -1
For 27:
Sine: (1-√3)/2
Cosine: (1+√3)/2
Tangent: -√3
For 29:
Sine: -(√3-1)/2
Cosine: (1+√3)/2
Tangent: -√3
For 16:
Sine: √3/2
Cosine: -1/2
Tangent: -√3
For 18:
Sine: -√2/2
Cosine: √2/2
Tangent: -1
For 20:
Sine: (√3-1)/2
Cosine: (1+√3)/2
Tangent: √3
For 22:
Sine: -(1+√3)/2
Cosine: (√3-1)/2
Tangent: -1/√3
For 24:
Sine: 1/2
Cosine: √3/2
Tangent: 1/√3
For 26:
Sine: √2/2
Cosine: -√2/2
Tangent: 1
For 28:
Sine: (√3+1)/2
Cosine: (1-√3)/2
Tangent: -1/√3
For 30:
Sine: -(1-√3)/2
Cosine: (√3+1)/2
Tangent: √3
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