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3 years ago

If Pentagon ABCDE was dilated to create Pentagon A'B'C'DE', what rule was used?

Mathematics

1 answer:

svetlana [45]3 years ago

5 0

Answer:

<em>B</em> $(x,y) \rightarrow \left(\frac{5}{2}x,\frac{5}{2}y\right)$

Step-by-step explanation:

<u>Dilations</u>

Given a point A(x,y) and a scale factor k the dilated image of A, called A' is calculated as A'=(kx,ky), assuming the same scale factor is applied in both axes.

The pentagon ABCDE was dilated to create pentagon A'B'C'D'E'. To find the dilaton rule used, we must find two clear points where the coordinates of both axes can be easily read from the graph.

Point C(-2,0) maps to C'(-5,0). This gives us the scale factor for the x-axis of -5/(-2)= 5/2.

The y-coordinate of E is 2 and the y-coordinate of E' is 5. This gives us the same scale factor for the y-axis of 5/2.

Thus, the rule to dilate the pentagon is:

B $\mathbf{(x,y) \rightarrow \left(\frac{5}{2}x,\frac{5}{2}y\right)}$

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jeyben [28]

Hi,

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Cost of the fence= a*6+a*12+b*12=3000 $.

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2 years ago

Write the quadratic equation whose roots are 3 and −3, and whose leading coefficient is 2

GREYUIT [131]

Answer:

y = 2(x+3)(x-3)

Step-by-step explanation:

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x 1 to n represents the roots

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2 years ago

Name four fractions between 5/6 and 7/8

Nataly_w [17]

Answer:

101/120, 102/120, 103/120, and 104/120.

Step-by-step explanation:

First find the LCM of 6 and 8:24

So the two given fractions are 20/24, and 21/24

Since the the fractions will become decimals with the denominator 24, choose another denominator: 48

So the two given fractions are 40/48 and 42/48.

Even this denominator isn't possible so continue with different denominators.

As I notice, that each multiple increases by 1...

20-21, 40-42, 60-63, 80-84, and 100-105.

100-105 is the only numerator that has 4 numbers in-between so, the four fractions between 5/6 and 7/8 are: 101/120, 102/120, 103/120, and 104/120.

And the 2 given fractions are 100/120 and 105/120.

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2 years ago

Read 2 more answers

How do you make x the subject of the formula when your formula is y= abx? Please can you show me how you worked it out if you an

GrogVix [38]

If you want to solve for x you have to get x alone on one side of the equals sign. In order to do that you have to get rid of the stuff on the same side by doing the opposite operation so that it cancels. What you do on one side you do on. the other side to keep the equation true.

y= abx

a and b are being multiplied by x so divide by ab to cancel them out (ab/ab is 1).

y= abx/ab (ab cancels)

now divide the other side by ab to keep the equation true.

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7 0

3 years ago

Can someone please write this as a single logarithm and show work please and thank you

Mazyrski [523]

Answer:

$log_b(w^3y^4)$

Step-by-step explanation:

To write the expression as a single logarithm, or condense it, use the properties of logarithms.

1) The power property of logarithms states that $log_ax^r = rlog_ax$ . In other words, the exponent within a logarithm can be brought out in front so it's multiplied by the logarithm. This means that the number in front of the logarithm can also be brought inside the logarithm as an exponent.

So, in this case, we can move the 3 and the 4 inside the logarithms as exponents. Apply this property as seen below:

$3log_bw+4log_by\\=log_bw^3+log_by^4$

2) The product property of logarithms states that $log_axy = log_ax+log_ay$ . In other words, the logarithm of a product is equal to the sum of the logarithms of its factors. So, in this case, write the expression as a single logarithm by taking the log (keep the same base) of the product of $w^3$ and $y^4$ . Apply the property as seen below and find the final answer.

$log_bw^3+log_by^4\\=log_b(w^3y^4)$

So, the answer is $log_b(w^3y^4)$ .

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2 years ago