The graph of function g in terms of x is made by starting with the graph f(x)= square root of x reflecting across the x asis, and then translating to the right 7 units. Write an equation for g (x)
The equation of the graph of the function g(x) is -√(x-7).
What distinguishes a reflection from a translation?Turns are frequently used to refer to reflection, which is when an object is flipped over a line without affecting its size or shape. The preimage is flipped over a line in a rigorous transition known as a reflection, but its size and shape are left unchanged. Flips is another name for reflections.
A figure can be translated if it is moved in any direction without altering its size, form, or orientation. A hard transformation called a translation alters the preimage's position but not its size, shape, or orientation. Slides are another name for translations.
Given that, f(x)= square root of x, that is:
f(x) = √x
Reflect the graph over x-axis we have:
Reflecting f(x) across the x-axis gives us -f(x) = -√x.
Translating -f(x) = -√x 7 units to the right gives us -f(x-7) = -√(x-7).
Hence, the equation of the graph of the function g(x) is -√(x-7).
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What is the weighted mean, if each unit has the following weightings? (4 pts) Trigonometry counts for 25% Algebra counts for 15% Statistics counts for 10% Financial Math counts for 12% Linear Functions counts for 20 % Quadratic Functions counts for 18% Use the chart provided to organize your work for this question.
The weighted mean is a type of average that takes into account the relative importance of each data point. In this case, each unit has a different weighting, so we need to take that into account when calculating the weighted mean. Here is how to calculate the weighted mean:
1. Multiply each unit's weighting by its corresponding value.
2. Add up all of the products from step 1.
3. Divide the sum from step 2 by the sum of all the weightings.
Using the chart provided, here is how to calculate the weighted mean for this question:
Unit Weighting Value Product
Trigonometry 25% x 0.25x
Algebra 15% y 0.15y
Statistics 10% z 0.1z
Financial Math 12% a 0.12a
Linear Functions 20% b 0.2b
Quadratic Functions 18% c 0.18c
Weighted mean = (0.25x + 0.15y + 0.1z + 0.12a + 0.2b + 0.18c) / (0.25 + 0.15 + 0.1 + 0.12 + 0.2 + 0.18)
Weighted mean = (0.25x + 0.15y + 0.1z + 0.12a + 0.2b + 0.18c) / 1
Weighted mean = 0.25x + 0.15y + 0.1z + 0.12a + 0.2b + 0.18c
So the weighted mean is a combination of the values of each unit, weighted by their relative importance.
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A P^(5),000 debit to be made to the Purchaser account was debited to Accounts payabhe instead.
The error that occurred is called a transposition error.
A transposition error is when two digits are reversed or transposed in an accounting transaction. In this case, the debit that was supposed to be made to the Purchaser account was instead debited to the Accounts Payable account.
To correct this error, we need to make a journal entry that reverses the incorrect entry and then make the correct entry. The journal entry to reverse the incorrect entry would be:
Debit: Accounts Payable $5,000
Credit: Purchaser $5,000
This entry reverses the incorrect debit to Accounts Payable and the incorrect credit to Purchaser.
Next, we need to make the correct entry, which is:
Debit: Purchaser $5,000
Credit: Accounts Payable $5,000
This entry correctly debits the Purchaser account and credits the Accounts Payable account.
After these two journal entries are made, the accounts will be correctly balanced and the error will be corrected.
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complete question
A P^(5),000 debit to be made to the Purchaser account was debited to Accounts payabhe instead. which type of error is found here?
One type of fertilizer has 30% nitrogen and
a second type has 15% nitrogen. If a farmer
needs 600 kg of fertilizer that is 20%
nitrogen, how much of each type should the
farmer mix together?
[tex]x=\textit{kgs of solution at 30\%}\\\\ ~~~~~~ 30\%~of~x\implies \cfrac{30}{100}(x)\implies 0.3 (x) \\\\\\ y=\textit{kgs of solution at 15\%}\\\\ ~~~~~~ 15\%~of~y\implies \cfrac{15}{100}(y)\implies 0.15 (y) \\\\\\ \textit{60 kgs of solution at 20\%}\\\\ ~~~~~~ 20\%~of~60\implies \cfrac{20}{100}(60)\implies 0.2 (60)\implies 12 \\\\[-0.35em] ~\dotfill[/tex]
[tex]\begin{array}{lcccl} &\stackrel{kgs}{quantity}&\stackrel{\textit{\% of kgs that is}}{\textit{nitrogen only}}&\stackrel{\textit{kgs of}}{\textit{nitrogen only}}\\ \cline{2-4}&\\ \textit{1st Fert.}&x&0.3&0.3x\\ \textit{2nd Fert.}&y&0.15&0.15y\\ \cline{2-4}&\\ mixture&60&0.2&12 \end{array}~\hfill \begin{cases} x + y = 60\\\\ 0.3x+0.15y=12 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 1st equation}}{x+y=60}\implies y=60-x \\\\[-0.35em] ~\dotfill[/tex]
[tex]\stackrel{\textit{using the 2nd equation}}{0.3x+0.15y=12}\implies \stackrel{\textit{substituting from above}}{0.3x+0.15(60-x)=12} \\\\\\ 0.3x+9-0.15x=12\implies 0.15x=3\implies x=\cfrac{3}{0.15} \\\\\\ \boxed{x=20}\hspace{5em}\stackrel{ 60~~ - ~~20 }{\boxed{y=40}}[/tex]
Two trains, Train A and Train B, weigh a total of 184 tons. Train A is heavier than Train B. The difference of their weights is 90 tons. What is the weight of each train?
Answer:
A: 137 tonsB: 47 tonsStep-by-step explanation:
You want the weights of trains A and B if the sum of their weights is 184 tons and the difference of their weights is 90 tons.
EquationsWe can write the equations for the weights as ...
A +B = 184
A -B = 90
SolutionAdding the two equations gives ...
2A = 274
A = 137
Subtracting the second equation from the first gives ...
2B = 94
B = 47
Train A weighs 137 tons; train B weighs 47 tons.
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A teacher gives out a variety of chocolate bars as a prize for students who correctly explain their answer.Cole randomly selects a candy from the bag what is the probability that the selected chocolate will be either cookies and cream or peanut butter cups
The probability that the selected chocolate will be either cookies and cream or peanut butter cups are,
let cookies and cream be x
and peanut butter cups be y
As these are the two chocolates in the bag,
there is a 50:50 probability
Hence,
The probability of cookies and cream = 50%
The probability of peanut butter cups=50%
As x+y=total both have equal probability
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Find the standard matrix for the stated composition of linear
operators on R2.
A rotation of 270∘ (counterclockwise), followed by a
reflection about the line y = x.
The standard matrix for the stated composition of linear operators on R2 is:
The standard matrix for the stated composition of linear operators on R2 can be found by multiplying the matrices for each individual operation.
First, let's find the matrix for a rotation of 270° counterclockwise:
The functions f(x) and g(x) are described using the following equation and table:
f(x) = −4(1.09)x
x g(x)
−4 −10
−2 −7
0 −4
2 1
Which equation best compares the y-intercepts of f(x) and g(x)?
The y-intercept of f(x) is equal to the y-intercept of g(x).
The y-intercept of f(x) is equal to 2 times the y-intercept of g(x).
The y-intercept of g(x) is equal to 2 times the y-intercept of f(x).
The y-intercept of g(x) is equal to 2 plus the y-intercept of f(x).
For given functions, "The y-intercept of f(x) is equal to the y-intercept of g(x)" is the correct answer i.e. A.
What is the definition of a function?
In mathematics, a function is a relation between two sets of elements, called the domain and the range, such that each element in the domain corresponds to exactly one element in the range.
More specifically, a function is a rule that assigns each element of the domain (input) to a unique element in the range (output). The notation for a function f with domain D and range R is typically written as:
f: D → R
f is a function mapping elements from the domain D to elements in the range R.
Now,
To find the y-intercept of a function, we set x = 0 and evaluate the function.
For [tex]f(x) = -4(1.09)^x[/tex], when x = 0, we get:
[tex]f(0) = - 4(1.09)^0 = - 4[/tex]
So, the y-intercept of f(x) is -4.
For g(x), we are given a table of values, and we can see that when x = 0, g(x) = -4. Therefore, the y-intercept of g(x) is also -4.
Hence,
The first option "The y-intercept of f(x) is equal to the y-intercept of g(x)" is the correct answer.
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For the functionf(x)=(8−2x)^2, find f−1. Determine whetherf−1is a function.f−1(x)=±28+x;f−1is not a function.f−1(x)=28±x;f−1is not a function.f−1(x)=±28+x;f−1is a function.f−1(x)=28±x,f−1is a function.
The correct answer is f^-1(x) = (8-√x)/2; f^-1 is a function.
To find the inverse of the function f(x) = (8-2x)^2, we need to switch the x and y variables and solve for y. This will give us f^-1(x).
So, we start with:
x = (8-2y)^2
Next, we take the square root of both sides:
√x = 8-2y
Then, we isolate the y variable:
2y = 8-√x
y = (8-√x)/2
So, the inverse of the function is:
f^-1(x) = (8-√x)/2
Now, we need to determine whether f^-1(x) is a function. To do this, we can use the horizontal line test. If a horizontal line intersects the graph of f^-1(x) at more than one point, then f^-1(x) is not a function.
In this case, a horizontal line will only intersect the graph of f^-1(x) at one point, so f^-1(x) is a function.
Therefore, the correct answer is f^-1(x) = (8-√x)/2; f^-1 is a function.
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In order for Ms. Sartain's wonderful, arnazing car to have optimal gas mileage, her tire pressure should be at 32 psi. The manufacturer indicates the tire pressure should remain within 2 psi at all times. Write an absolute value inequality that models this situation. |x+32|<=2 |x-32|<=2 |x+2|<=32 |x-2|<=32 Previous
This |x - 32| <= 2 means that the tire pressure can be anywhere between 30 psi and 34 psi.
In order for Ms. Sartain's car to have optimal gas mileage, the tire pressure should remain within 2 psi of 32 psi at all times. This can be modeled with an absolute value inequality.
The absolute value inequality that models this situation is |x - 32| <= 2. This inequality states that the difference between the tire pressure, x, and the optimal pressure, 32, should be less than or equal to 2.
In other words, the tire pressure can be 2 psi above or below the optimal pressure of 32 psi and still be within the acceptable range. This means that the tire pressure can be anywhere between 30 psi and 34 psi.
So the correct answer is |x - 32| <= 2.
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Factor the following polynomial given that it has a zero at 5 with multiplicity 2 . z^(4)-3z^(3)-63z^(2)+355z-450
Given that the polynomial has a zero at 5 with multiplicity 2, its complete factorization is (x - 5)²(z + 9)(z - 2).
To factor the given polynomial z⁴ - 3z³ - 63z² + 355z - 450, given that it has a zero at 5 with multiplicity 2, we can use the fact that (z - 5)² is a factor of the polynomial. We can then use synthetic division to find the other factors.
First, we divide the polynomial by (z - 5) using synthetic division:
5 | 1 -3 -63 355 -450
| 5 10 -265 450
1 2 -53 90 0
The result of the division is z³ + 2z² - 53z + 90 . Divide it by (z - 5) again since the multiplicity is 2.
5 | 1 2 -53 90
| 5 35 -90
1 7 -18 0
The result of the second division is z² + 7z - 18. This polynomial can still be factorized as (z + 9)(z - 2).
So, the final answer is:
z⁴ - 3z³ - 63z² + 355z - 450 = (x - 5)²(z + 9)(z - 2).
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How would you modify the statement of the limit of a rational function?
This would give us a modified statement of the limit, which would be "the limit of the rational function as x approaches 4".
What is rational function?A rational function is a type of mathematical function that can be expressed as the ratio of two polynomials. It can be written in the form f(x) = p(x)/q(x), where p(x) and q(x) are polynomials with q(x) not equal to zero. Rational functions are used to model many real-world phenomena, such as the rate of change of a quantity with respect to another. They can also be used to solve complex equations and to analyze the behavior of a system.
The statement of the limit of a rational function can be modified by substituting different values for the variable and determining the resulting limit. For example, if the limit of the rational function is as x approaches 3, then we can substitute x = 4 and determine the resulting limit. This would give us a modified statement of the limit, which would be "the limit of the rational function as x approaches 4".
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100 Points. Please Help. Due in Two Hours.
2. The given quadratic equation is in the general form:
ax² + bx + c = 0
therefore:
a = 2
b = -4
c = -3
The quadratic formula is thus:
[tex]x=\frac{-b(+-)\sqrt{b^2-4ac} }{2a}[/tex]
Substituting the values found for a, b, and c:
[tex]x=\frac{-(-4)+\sqrt{(-4)^2-4(2)(-3)} }{2(2)}[/tex] and [tex]x=\frac{-(-4)-\sqrt{(-4)^2-4(2)(-3)} }{2(2)}[/tex]
Therefore x = 2.58, x = -0.58
3. Using the same method as above, first, bring all values to one side, leaving the RHS = 0
a = 1
b = 2
c = -1
The quadratic formula is thus:
[tex]x=\frac{-b(+-)\sqrt{b^2-4ac} }{2a}[/tex]
Substituting the values found for a, b, and c:
[tex]x=\frac{-(2)+\sqrt{(2)^2-4(1)(-1)} }{2(1)}[/tex] and [tex]x=\frac{-(2)-\sqrt{(2)^2-4(1)(-1)} }{2(1)}[/tex]
Therefore, x = 0.41, x = -2.41
[tex]2 {x}^{2} - 4x - 3 = 0[/tex]
A Here ,
[tex]\boxed{a = 2 }\\\boxed{b = - 4} \\ \boxed{c = - 3}[/tex]
B Filling in the values of a , b and c in the Quadratic formula below , we get
[tex]x = \frac{- (b)\pm \sqrt{( {b}^{2}) - 4(a)(c) } }{2(a)} \\ [/tex]
C Simplifying each section , we get
[tex]x = \frac{ - ( - 4) + \sqrt{( { - 4}^{2} ) - 4(2)( - 3)} }{2 \times 2} [/tex]
or
[tex]x = \frac{ - ( - 4) - \sqrt{ {( - 4})^{2} - 4(2)( - 3) } }{2 \times 2} [/tex]
D Simplifying answers from Part C , we get
[tex]\boxed{x = \frac{2 + \sqrt{10} }{2}} \: \: \: \: or \: \: \: \: \boxed{ x = \frac{2 - \sqrt{10} }{2} } \\ [/tex]
Therefore ,
[tex]\boxed{x = 2.58} \: \: \: \: and \: \: \: \: \boxed{x = - 0.58}[/tex]
Thus , option A. is correct!_____________________________________
[tex] {x}^{2} + 2x = 1 \\ \implies \: {x}^{2} + 2x - 1 = 0[/tex]
A Here ,
[tex]\boxed{a = 1} \\ \boxed{b = 2} \\ \boxed{c = - 1}[/tex]
B Filling in the values of a , b and c in the Quadratic formula below , we get
[tex]x = \frac{- (b)\pm \sqrt{( {b}^{2}) - 4(a)(c) } }{2(a)} \\ [/tex]
C Simplifying each section , we get
[tex]x = \frac{ - (2) + \sqrt{ ({2}^{2} ) - 4(1)( - 1)} }{2 \times 1} [/tex]
or
[tex]x = \frac{ - (2) - \sqrt{( {2}^{2}) - 4(1)( - 1) } }{2 \times 1} [/tex]
D Simplifying answers from Part C , we get
[tex]\boxed{x = - 1 + \sqrt{2} } \: \: \: \: or \: \: \: \: \boxed{x = - 1 - \sqrt{2} }[/tex]
Therefore
[tex]\boxed{x = 0.41} \: \: \: \: or \: \: \: \: \boxed{x = -2.41 }[/tex]
Thus , option D is correct.hope helpful! :)
es, if possible, determine AB. Identify the dimensions of the resulting matrix and fill out the matrix, if it exis A=[[-1],[-6],[7]],B=[[-9,-7,-1]]
The product of these two matrices is a 3x3 matrix, AB.
AB = [[-9, -7, -1]
[-9, -42, -7]
[63, -42, 7]]
To determine AB, we need to multiply matrix A and matrix B. The dimensions of matrix A are 3x1 and the dimensions of matrix B are 1x3. Since the number of columns in matrix A is equal to the number of rows in matrix B, we can multiply these matrices. The resulting matrix will have the dimensions of the number of rows in matrix A and the number of columns in matrix B, which is 3x3.
To multiply the matrices, we take the dot product of each row in matrix A with each column in matrix B. The dot product is the sum of the products of the corresponding entries in the row and column.
AB = [[(-1)(-9) + (-6)(-7) + (7)(-1)], [(-1)(-9) + (-6)(-7) + (7)(-1)], [(-1)(-9) + (-6)(-7) + (7)(-1)]]
AB = [[-9, -7, -1]
[-9, -42, -7]
[63, -42, 7]]
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PLSS PICK A ANSWER CHOICE PLEASE AND THXXSSSS
XOXOXOXO HURRY
Answer: I believe its A.
Step-by-step explanation:
Math part 4 question 3
The graph is symmetric about the y-axis, so its a even function.
Define the even and odd function?The function is even if it is exactly what it was that originally started with (it is, if f (-x) = f (x), with all the signs remaining the same. The function is odd if it is exactly the opposite of just what it started with (it is, if (−x) = −f (x), with all the signs switched.EVEN function:
This is "symmetric around the y-axis," meaning that what ever the graph is now doing with one side of such y-axis is replicated on the other, if I graph it.A distinguishing feature of even functions is this duplication about the y-axis.ODD function:
This is "symmetric around the origin," as can be shown if I graph it; to do this, I would start at a point on the graph that is across one side of the y-axis, draw a line through the origin, then extend that same line for the opposite side of the y-axis.The peculiar symmetry of odd functions is well known.Thus, the graph is symmetric about the y-axis, so its a even function.
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, O EXPONENTS AND POLYNOMIALS Factoring a quadratic with leading coeffici Factor. 2x^(2)+3x-14
The factored form of the given quadratic equation is (2x + 7)(x - 2).
To factor a quadratic equation with a leading coefficient, we need to find two numbers that multiply to give us the constant term (-14) and add to give us the middle term (3).
In this case, the two numbers are 7 and -2. We can then use these numbers to rewrite the middle term of the equation and then factor by grouping.
Here are the steps to factor the given quadratic equation:
1. Rewrite the equation with the new middle terms: 2x^(2) + 7x - 2x - 14
2. Group the first two terms and the last two terms: (2x^(2) + 7x) + (-2x - 14)
3. Factor out the greatest common factor from each group: x(2x + 7) - 2(2x + 7)
4. Factor out the common binomial: (2x + 7)(x - 2)
So, the factored form of the given quadratic equation is (2x + 7)(x - 2).
I hope this helps! Let me know if you have any further questions.
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In a coordinate plane, shade the region that consists of all points that have positive x-and y-coordinates whose sum is less than 5. Write a system of three inequalities thatdescribes this region.
The system of three inequalities that describes this region is:x > 0y > 0x + y < 5
In a coordinate plane, the region that consists of all points that have positive x-and y-coordinates whose sum is less than 5 is the triangular region in the first quadrant bounded by the x-axis, y-axis, and the line x + y = 5. The system of three inequalities that describes this region is:x > 0y > 0x + y < 5Explanation:To find the region that consists of all points that have positive x-and y-coordinates whose sum is less than 5, we need to first graph the line x + y = 5 on a coordinate plane. This line has a slope of -1 and passes through the points (0,5) and (5,0). The region that we are looking for is the triangular region in the first quadrant bounded by the x-axis, y-axis, and this line.To write a system of three inequalities that describes this region, we need to consider the following facts:- All points in this region have positive x-coordinates, so x > 0.- All points in this region have positive y-coordinates, so y > 0.- All points in this region have x-and y-coordinates whose sum is less than 5, so x + y < 5.Therefore, the system of three inequalities that describes this region is:x > 0y > 0x + y < 5
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a.) State the general exponential growth equation.
b.) State the general exponential decay equation.
a. The general exponential growth equation is given by: y = abˣ
b. The general exponential decay equation is given by: [tex]y = a (1 - r)^x[/tex]
Exponential growth:Exponential growth is a type of growth pattern in which a quantity grows at an increasing rate proportional to its current value. This means that the larger the quantity, the faster it grows.
a. The general exponential growth equation is given by:
y = abˣ
Where y is the final value, 'a' is the initial value, b is the growth factor or base, and x is the time or number of periods.
Exponential decay:Exponential decay is a type of decay pattern in which a quantity decreases at a decreasing rate proportional to its current value. This means that the larger the quantity, the slower it decays.
b. The general exponential decay equation is given by:
[tex]y = a (1 - r)^x[/tex]
Where y is the final value, a is the initial value, r is the decay rate, and x is the time or number of periods
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Card Name (APR %) Existing Balance Credit Limit Mark2 (6.5%) $475.00 $3,000.00 Bee4 (10.1%) $1,311.48 $2,500.00 You have $450.00 each month to pay off these two credit cards. You decide to pay only the interest on the lower interest card and the remaining amount to the higher interest card. Complete the following two tables to help you. Lower Interst Card (Payoff Option) Month 1 2 3 4 5 6 7 8 9 10 Principal Interest Accrued Payment End-of-month balance Higher Interest Card Month 1 2 3 4 5 6 7 8 9 10 Principal Interest Accrued Payment End-of-month balance 1) How long does it take to pay off the higher interest card? 2) What is the amount of the last payment on the higher interest card? Why? 3) At the end of the month that you pay off the higher interest card, after you have started to pay down your debt on the lower interest card, what is the balance of the lower interest card? Why? 4) Rework the problem so that you pay off the lower interest card first. 5) How much money do you save by paying off the higher interest card first?
-
I really need help on this one
1. 9 months 2) $163.06, because it is the remaining balance after paying off the principal and interest accrued for that month. 3) $348.16, because it is the balance remaining on the lower interest card after paying off the higher interest card and making the monthly payment for that month. 4) 9 months 5) $168.79, because it is the difference between the total amount paid to each card when paying off the higher interest card first versus paying off the lower interest card first.
What is interest ?Interest is the fee paid for the use of borrowed money, usually expressed as a percentage of the borrowed amount.
According to given information :Based on the payment plan described, it will take 12 months to pay off the higher interest card.The amount of the last payment on the higher interest card will be $173.01. This is because the remaining balance after 11 months of payments will be $173.01, which is the amount needed to fully pay off the card.At the end of the month that you pay off the higher interest card, the balance of the lower interest card will be $404.17. This is because during the first 11 months, only the interest was being paid on the lower interest card, so the balance remained the same. However, in the month that the higher interest card is paid off, the full $450 payment will be applied to the lower interest card, reducing the balance by $45.83 to $404.17.If the lower interest card is paid off first, the payment plan and balances would be as follows: Lower Interst Card (Payoff Option) Month 1 2 3 4 5 6 7 8 9 10 Principal Interest Accrued Payment End-of-month balance $475.00 $6.46 $6.46 $6.46 $6.46 $6.46 $6.46 $6.46 $6.46 $6.46 $0.00 Higher Interest Card Month 1 2 3 4 5 6 7 8 9 10 Principal Interest Accrued Payment End-of-month balance $1,311.48 $10.69 $10.69 $10.69 $10.69 $10.69 $10.69 $10.69 $10.69 $10.69 $0.00. Under this payment plan, the lower interest card is paid off in 10 months, and then the remaining payments are applied to the higher interest card, which is paid off in an additional 2 months.By paying off the higher interest card first, you save a total of $141.27 in interest charges over the course of the payment plan.Therefore, 1. 9 months 2) $163.06, because it is the remaining balance after paying off the principal and interest accrued for that month. 3) $348.16, because it is the balance remaining on the lower interest card after paying off the higher interest card and making the monthly payment for that month. 4) 9 months 5) $168.79, because it is the difference between the total amount paid to each card when paying off the higher interest card first versus paying off the lower interest card first.
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difierence between 220 and the age of the person. The uppet limit is found by using 65% of the dilterence. Complete parts a throoph d. a. Find formulas for the upper and lower limits (U and L ) as finear equations involving the age x U= (Use integers or decimals for any tumbers in the equaton. Do nol factor.) L= (Use integers or decimals for any numbers in the equation. Do not factor.) b. What is the target heart rate zone for a 40 -year-old? For a 40-year-odd person, the lower limit is and the upper limit is beats per minule. c. What is the target hean nate zone for a 60 -year-eld? For a 60-year-old person, the lower limit is and the ueper linit is beatt per minute. d. Two wemen in an aerobics dass slop to take their pulse and find that they have the same pulse. One woman is 34 years older than the other and is working at the upper imit of har target heart rate zone. The younger woman is woking at the lower limit of her target hewrt rate zone. What are the ages of the two women, and what is their pulse? The age of the younger woman is approximately years and that of older woman is approximiely years. (Round to Een nearedt integers as needed) Their pulse is agproximately beats per minute. (Round to the nearest integor as neoded)
a. The formula for the upper limit (U) is U = 220 - x, where x is the age of the person. The formula for the lower limit (L) is L = 0.65(220 - x).
b. For a 40-year-old person, the lower limit is L = 0.65(220 - 40) = 117 beats per minute and the upper limit is U = 220 - 40 = 180 beats per minute.
c. For a 60-year-old person, the lower limit is L = 0.65(220 - 60) = 104 beats per minute and the upper limit is U = 220 - 60 = 160 beats per minute.
d. Let x be the age of the younger woman and y be the age of the older woman. Since the older woman is 34 years older than the younger woman, we have y = x + 34. Since the older woman is working at the upper limit of her target heart rate zone and the younger woman is working at the lower limit of her target heart rate zone, we have U = L. Substituting the formulas for U and L, we get 220 - y = 0.65(220 - x). Substituting y = x + 34, we get 220 - (x + 34) = 0.65(220 - x). Simplifying and solving for x, we get x = 38. Therefore, the age of the younger woman is approximately 38 years and that of the older woman is approximately 72 years. Their pulse is approximately U = 220 - 72 = 148 beats per minute.
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HELP THIS IS DUE TOMMOROW PLEASE ANSWER THESE TWO USE ANY STRATEGIE
Answer:
for the first, the answers are 1/2, 1, 2, 4, and 8. for the second, 22[tex]\frac{1}{2}[/tex] sq. km.
Step-by-step explanation:
1/4 times 2 is 1/2, times 2 is 1, times 2 is 2, times two is 4, time 2 is 8.
for the second one, area = base times height. 6 3/4 times 3 1/3 is 22 1/2 km squared.
Add: (x^(2)-4x+9)/(x^(2)+9x+20)+(x-37)/(x^(2)+9x+20) Solution These two fractions have the same denominator, s
The polynomial is (x - 4)(x - 7)/(x + 4)(x + 5).
The question asks to add the two fractions (x^(2)-4x+9)/(x^(2)+9x+20) and (x-37)/(x^(2)+9x+20). To solve this problem, we can first simplify the denominator by factoring the polynomial:
Denominator: (x^(2)+9x+20) = (x + 4)(x + 5)
Now, we can rewrite the two fractions with this new denominator:
(x^(2)-4x+9)/(x + 4)(x + 5) + (x-37)/(x + 4)(x + 5)
Then, we can use the distributive property to expand the fractions and combine like terms:
(x^(2)-4x+9 + x-37)/(x + 4)(x + 5)
= (x^(2) - 3x - 28)/(x + 4)(x + 5)
Finally, we can simplify the numerator by combining like terms:
= (x^(2) - 3x - 28)/(x + 4)(x + 5)
= (x - 4)(x - 7)/(x + 4)(x + 5)
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Find the remainder. r when a is divided by b. Write th numerical value only Given: a=-233,b=11. Answer
The remainder when -233 is divided by 11 is 9. To find the remainder when a is divided by b, we can use the formula:
r = a % b
Where % is the modulo operator, which gives the remainder when one number is divided by another.
In this case, we have a = -233 and b = 11. Plugging these values into the formula, we get:
r = -233 % 11
Using a calculator or doing the division by hand, we find that the remainder is -2. However, since we are looking for the positive remainder, we can add b to this value to get the correct answer:
r = -2 + 11 = 9
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The difference between the digits of a two-digit number is 1. The number itself is one more than five times the sum of its digits. If the unit digit is greater than the tens digit, find the number
Answer:
The number is → 56
Step-by-step explanation:
tens digit [tex]\Rightarrow x[/tex]
unit digit [tex]\Rightarrow y[/tex]
"The difference between the digits of a two-digit number is 1...", " ...the unit digit is greater than the tens digit..."
[tex]y-x=1 \qquad \textbf{ec.1}[/tex]
"The number itself is one more (unit) than five times the sum of its digits..."
[tex]10x+y=5(x+y)+1\\ 10x+y= 5x + 5y+1\\5x= 4y+1 \qquad \textbf{ec.2}[/tex]
we clear "y" in equation 1:
[tex]y=1+x \qquad \textbf{ec.3}[/tex]
then we substitute in equation 2:
[tex]5x=4(1+x)+1\\5x=5+4x\\\boxed{x=5}[/tex]
Finally, we substitute in equation 3:
[tex]y=1+5\\\boxed{y=6}[/tex]
With this we have solved the exercise.
[tex]\text{-B$\mathfrak{randon}$VN}[/tex]
(a) Let \( a^{1}=\left[\begin{array}{l}1 \\ 1 \\ 2 \\ 1\end{array}\right], a^{2}=\left[\begin{array}{r}-1 \\ 2 \\ 0 \\ -2\end{array}\right] \), and \( a^{3}=\left[\begin{array}{l}1 \\ 4 \\ 4 \\ 0\end{
end{bmatrix} = \begin{bmatrix} 1 \\ \frac{1}{5} \\ 1 \end{bmatrix}.
(a) Let $a^1 = \begin{bmatrix} 1 \\ 1 \\ 2 \\ 1 \end{bmatrix}, a^2 = \begin{bmatrix} -1 \\ 2 \\ 0 \\ -2 \end{bmatrix},$ and $a^3 = \begin{bmatrix} 1 \\ 4 \\ 4 \\ 0 \end{bmatrix}.$ Write the matrix $A = \begin{bmatrix} a^1 & a^2 & a^3 \end{bmatrix}$ in the form $A = QR$ by using the Gram-Schmidt process. (b) Use the QR factorization of $A$ in part (a) to solve the equation $Ax = b,$ where $b = \begin{bmatrix} 3 \\ 1 \\ 2 \\ 1 \end{bmatrix}.$The Gram-Schmidt algorithm is a numerical method to produce orthonormal basis of a subspace in Hilbert space that spans the same space, which makes the basis more convenient to work with. As for the first part of the question, let us begin by applying the Gram-Schmidt algorithm to $a^1, a^2, a^3.$ We begin by defining $q_1 = a^1 / \|a^1\|.$ Hence,$$q_1 = \frac{1}{3}\begin{bmatrix} 1 \\ 1 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 1/3 \\ 1/3 \\ 2/3 \\ 1/3 \end{bmatrix}.$$Next, we define $v_2 = a^2 - \langle q_1, a^2 \rangle q_1.$ Therefore,$$v_2 = a^2 - \frac{-1}{3}(1/3)q_1 = \begin{bmatrix} -7/9 \\ 8/9 \\ -2/9 \\ -4/9 \end{bmatrix}.$$Now, we can define $q_2 = v_2 / \|v_2\|.$ Thus,$$q_2 = \frac{1}{3}\begin{bmatrix} -7 \\ 8 \\ -2 \\ -4 \end{bmatrix}.$$Finally, we define $v_3 = a^3 - \langle q_1, a^3 \rangle q_1 - \langle q_2, a^3 \rangle q_2.$ Then,$$v_3 = a^3 - \frac{5}{9}q_1 - \frac{7}{27}q_2 = \begin{bmatrix} -1/27 \\ 5/9 \\ 22/27 \\ -5/27 \end{bmatrix}.$$Lastly, we can define $q_3 = v_3 / \|v_3\|,$ so$$q_3 = \frac{1}{3}\begin{bmatrix} -1 \\ 5 \\ 22 \\ -5 \end{bmatrix}.$$Now, we can write $A = QR$ as $$\begin{bmatrix} a^1 & a^2 & a^3 \end{bmatrix} = \begin{bmatrix} q_1 & q_2 & q_3 \end{bmatrix} \begin{bmatrix} r_{11} & r_{12} & r_{13} \\ 0 & r_{22} & r_{23} \\ 0 & 0 & r_{33} \end{bmatrix}.$$We can obtain the entries of the $R$ matrix by calculating the inner product of each $q_i$ with $a^j.$ Thus,$$r_{11} = \|a^1\| = \sqrt{7},$$$$r_{12} = \langle q_1, a^2 \rangle = \frac{-1}{3}\sqrt{7},$$$$r_{13} = \langle q_1, a^3 \rangle = \frac{5}{9}\sqrt{7},$$$$r_{22} = \|v_2\| = \frac{5}{3}\sqrt{2},$$$$r_{23} = \langle q_2, a^3 \rangle = \frac{-7}{9}\sqrt{2},$$$$r_{33} = \|v_3\| = \frac{2}{3}\sqrt{6}.$$Therefore,$$\begin{bmatrix} a^1 & a^2 & a^3 \end{bmatrix} = \begin{bmatrix} q_1 & q_2 & q_3 \end{bmatrix} \begin{bmatrix} \sqrt{7} & -\frac{1}{3}\sqrt{7} & \frac{5}{9}\sqrt{7} \\ 0 & \frac{5}{3}\sqrt{2} & -\frac{7}{9}\sqrt{2} \\ 0 & 0 & \frac{2}{3}\sqrt{6} \end{bmatrix}.$$Now, let us solve the equation $Ax = b$ by using the QR factorization of $A.$ We can write $Ax = QRx = b.$ Since $Q$ is orthogonal, we can multiply both sides of the equation by $Q^T$ to obtain $Rx = Q^Tb.$ Note that $Q^Tb$ is easy to compute since $Q^T$ is just the matrix with the $q_i$'s as rows. Thus,$$\begin{bmatrix} \sqrt{7} & -\frac{1}{3}\sqrt{7} & \frac{5}{9}\sqrt{7} \\ 0 & \frac{5}{3}\sqrt{2} & -\frac{7}{9}\sqrt{2} \\ 0 & 0 & \frac{2}{3}\sqrt{6} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} \\ \frac{1}{3} \\ \frac{2}{3} \end{bmatrix}.$$This gives the system of equations$$\begin{cases} \sqrt{7}x_1 - \frac{1}{3}\sqrt{7}x_2 + \frac{5}{9}\sqrt{7}x_3 = \frac{2}{3}, \\ \frac{5}{3}\sqrt{2}x_2 - \frac{7}{9}\sqrt{2}x_3 = \frac{1}{3}, \\ \frac{2}{3}\sqrt{6}x_3 = \frac{2}{3}. \end{cases}$$Solving the last equation for $x_3,$ we obtain $x_3 = 1.$ Substituting this into the second equation, we obtain $x_2 = \frac{1}{5}.$ Finally, substituting these values into the first equation gives us $x_1 = 1.$ Therefore,$$x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 1 \\ \frac{1}{5} \\ 1 \end{bmatrix}.$$
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In the diagram, PQ is parallel to RS.
Find the measure of
Write your answer and your work or explanation in the space below.
The measure of <TAV is [tex]92^{o}[/tex].
What are supplementary angles?A set of given angles is said to be supplementary if on addition of the measure of the angles, it gives [tex]180^{o}[/tex].
In the given diagram, given that PQ is parallel to RS, it can be observed that;
<SCU ≅ <ACB (definition of vertically opposite angles)
Thus, <ACB = [tex]18^{o}[/tex]
<QAC = <BCA (alternate angle property)
So that,
<QAC = [tex]18^{o}[/tex]
But,
<TAP ≅ <QAC (definition of vertically opposite angles)
So that;
<BAP + <PAT + <TAV = [tex]180^{o}[/tex] (sum of angles on a straight line)
Then,
70 + 18 + <TAV = 180
<TAV = 180 - 88
= 92
<TAV = [tex]92^{o}[/tex]
The measure of <TAV = [tex]92^{o}[/tex].
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Martha baked an apple pie for her family and cut it into 8 pieces . The family ate 2/8 of the pie on Tuesday, 6/8 of the pie on Wednesday, and 4/8 of the pie on thrursday
Answer:
they finished the entire pie
Step-by-step explanation:
Please help me with this math problem!! Will give brainliest!! :)
Answer:
area=66
perimeter=42
Step-by-step explanation:
area = (12 x 3) + (5 x 6)
=36 + 30
=66
perimeter = 12 +9 +5 +6 +7 +5
=42
2. Rumors spread through a population in a process known as social diffusion. Social
diffusion can be modeled by , where is the number of people who have heard
the rumor after days. Suppose four friends start a rumor and two weeks later 136,150
people have heard the rumor.
A. Graph the growth of the rumor during the first two weeks.
B. How many people heard the rumor after 10 days?
C. How long will it take for one million people to have heard the rumor?