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Elenna [48]
3 years ago
15

Determine whether the given scale factor is a reduction or an

Mathematics
2 answers:
Margaret [11]3 years ago
7 0

Answers:

  • Enlargement
  • The new larger rectangle is 16.8 inches by 8.4 inches

==============================================================

Explanation:

I'm assuming 7/5 is the scale factor. If so, notice how 7/5 = 1.4 which is larger than 1.

The scale factor being larger than 1 means we have an enlargement.

The image is going to be larger compared to the preimage.

The preimage rectangle is 12 in by 6 in. Multiply each dimension by 1.4 to get 12*1.4 = 16.8 and 6*1.4 = 8.4

The image rectangle is 16.8 inches by 8.4 inches

ehidna [41]3 years ago
5 0

Answer:

This is an enlargement. The dimensions of the scaled rectangle is 16.8 inches by 8.4 inches.

Step-by-step explanation:

This is a dilation.

A dilation makes a figure larger or smaller.

In this case, the scale factor is an enlargement; the figure will be larger.

The figure will be larger because the scale factor, 7/5, is larger than one (1).

If the scale factor were smaller than one (1), it would be a reduction.

In dilations, a scale factor is how much larger the image will be.

In other words, multiply the side lengths by the scale factor.

The original rectangle has a length of 12 inches and a width of 6 inches.

12*7/5=84/5

6*7/5=42/5

It will be more convenient to change the products, which are fractions, to decimals.

84/5=16.8

42/5=8.4

Therefore, the dimensions of the scale rectangle is 16.8 inches by 8.4 inches.

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Answer:

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Step-by-step explanation:

6r+3=4r-11

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Find the average value of the function on the given interval.<br> f(x)=(2x-1)1/2; [1,13]
Leni [432]

Answer:

The average value of the function on the given interval 6.5.

Step-by-step explanation:

Consider the given function is

f(x)=\dfrac{2x-1}{2}

We need to find the average value of the function on the given interval [1,13].

f(x)=\dfrac{2x}{2}-\dfrac{1}{2}

f(x)=x-0.5

The average value of the function f(x) on [a,b] is

Average=\dfrac{1}{b-a}\int\limits^b_a {f(x)} \, dx

Average value of the function on the given interval [1,13] is

Average=\dfrac{1}{13-1}\int\limits^{13}_{1} {x-0.5} \, dx

Average=\dfrac{1}{12}[\dfrac{x^2}{2}-0.5x]^{13}_{1}

Average=\dfrac{1}{12}[\dfrac{(13)^2}{2}-0.5(13)-(\dfrac{(1)^2}{2}-0.5(1))]

Average=\dfrac{1}{12}[78-0]

Average=6.5

Therefore, the average value of the function on the given interval 6.5.

6 0
3 years ago
(2x3)over 2-5 over 2
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Answer:

2x3/2-5/2

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6 divide by 2 is +3 and +3 divide by - 3 equals - 1

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3 years ago
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