An ant crawls 12 feet in 10 minutes. How far can it crawl in 20 minutes and in 30 minutes?

Answer:

x=5, y=-15

Step-by-step explanation:

7x+2y=5

3x+2y=-15

4x=20

 x=5

7(5)+2y=5

       2y=-30

         y=-15

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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The possible rational zeros of h(x) are 1, -1, 3, -3, 1/2, -1/2, 3/2, and -3/2.

According 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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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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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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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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To create a graph and write an equation to represent the list of ordered pairs given, we need to plot each point on a coordinate system.

x | y

--|--

2 |-3

1 |-1

-1| 3

3 |-5

Plotting these points on a coordinate plane, we get:

![Graph of ordered pairs]

To write an equation to represent these ordered pairs in the form y = mx + b, we need to find the slope of the line and the y-intercept.

To find the slope:

m = (y2 - y1) / (x2 - x1)

Let's use the points (1, -1) and (3, -5) to find the slope:

m = (-5 - (-1)) / (3 - 1)

m = -4 / 2

m = -2

Now that we know the slope, we can use any point and the slope to find the y-intercept (b).

Using the point (1, -1):

y = mx + b

-1 = (-2)(1) + b

b = 1

So the equation that represents these ordered pairs in the form y = mx + b is:

y = -2x + 1

And the graph of this equation looks like:

![Graph of equation y = -2x + 1]

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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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.

The 42 grams of aluminum implies that the chemist has 0.06429ml of aluminum

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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The percentage of workers who likes to walk or bike to work is P = 36 %

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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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]

An estimate for the mean age of the 50 people on the bus is 34 years.

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 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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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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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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The rate of change of f(x) = 2(2) is equal to the rate of change of the function in the graph.

A function is a relation between two sets of values in mathematics. It is a mathematical process that takes an input and produces an output. It is represented as an equation which describes the relationship between the input and the output. A function can also be used to represent a mathematical rule or procedure that takes a set of inputs and produces a set of outputs.

The rate of change, or slope, is the measure of how quickly one variable changes when the other changes. In the graph, the rate of change is a constant, meaning that for every one unit the x-value increases the y-value increases by two. This is the same as the rate of change of f(x) = 2(2), as for every two units the x-value increases the y-value increases by four. Both the graph and the equation of the function have a constant rate of change which is equal to two, so the rate of change of f(x) = 2(2) is equal to the rate of change of the function in the graph.

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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

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. :)

Answer:

You must get a 89%

Step-by-step explanation:

81+89=170

170/2= 85%

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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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 ]

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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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.

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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The table can be filled up accordingly:

1.  5 years compounded semi-annually:

b = (1 + 0.06/2)

y= 1800 (1 + 0.03)^10

x = 2

2. For the 5 years compounded quarterly:

b =  (1 + 0.06/4)

y =  1800 (1 + 0.015)^20

x =4

3. For the 5 years compounded monthly:

b = (1 + 0.06/12)

y = 1800 (1.005)^60

x = 12

4. For 10 years compounded annually:

b = (1 + 0.06/1)

x =  1800 (1.06)^10

y = 1

5. For 10 years compounded quarterly:

b =  (1 + 0.06/4)

y =1800 (1 + 0.005)^60

x = 4

6. For 10 years compounded monthly:

b = (1 + 0.06/12)

y = 1800 (1 + 0.005)^120

x = 12

To solve the compound interest we use the formula:

P(1+r/n)^(n*t).

For the 5 years compounded semi-annually:

A = 1800 (1 + 0.06/2)^2*5

A = 1800 (1 + 0.03)^10

A = 1800 (1.03)^10

A = 1800 (1.344)

A = 2419

For the 5 years compounded quarterly:

A = 1800 (1 + 0.06/4)^4*5

A = 1800 (1 + 0.015)^20

A = 1800 (1.015)^20

A= 1800 (1.3468)

A= 2424.24

For the 5 years compounded monthly:

A = 1800 (1 + 0.06/12)^12*5

A = 1800 (1 + 0.005)^60

A = 1800 (1.005)^60

A = 1800 (1.34885)

A = 2427.93

For 10 years compounded annually:

A= 1800 (1 + 0.06/1)^10

A = 1800 (1.06)^10

A = 1800 (1.7908)

A = 3223.53

For 10 years compounded semi-annually:

A = 1800 (1 + 0.06/2)^20

A = 1800 (1 + 0.03)^20

A = 1800 (1.03)^20

A =1800 (1.806)

A = 3251

For 10 years compounded quarterly:

A= 1800 (1 + 0.06/4)^4*10

A = 1800 (1 + 0.015)^40

A= 1800 (1.015)^40

A = 1800 (1.814)

A = 3265.23

For 10 years compounded monthly:

A= 1800 (1 + 0.06/12)^120

A = 1800 (1 + 0.005)^120

A = 1800 (1.005)^20

A= 1800 (1.8194)

A = 3274.9

The best pay period is that with the highest returns which is 10 years compounded monthly.

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The annual interest rate is 6.5%

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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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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According to the information, the values that complete the table are: 4, 6, and 150.

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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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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