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3 years ago

What scale factor was applied to the first rectangle to get the resulting image?

Mathematics

1 answer:

yulyashka [42]3 years ago

7 0

You would take 3 into 7.5 to get that scale factor.

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Find the midpoint of the line segment joining points A and B.<br><br><br>A: (8,-7)<br><br>B: (2,1)

egoroff_w [7]

Answer:

8+2/2

= 5

-7+1 / 2

= -3

(5,-3) is the correct answer

4 0

3 years ago

The ratio of the ages of Mandy and Sandy is 2:5. After 8 years, their ages will be in the ratio 1:2. What is the difference betw

swat32

A ratio shows us the number of times a number contains another number. Caitlin is right.

<h3>What is a Ratio?</h3>

A ratio shows us the number of times a number contains another number.

The ratio of the ages of Mandy and Sandy is 2:5. Therefore, the ratio will be,

M/S = 2/5

M = 2S/5

After 8 years, their ages will be in the ratio 1:2, therefore,

(M+8)/(S+8) = 1/2

2M + 16 = S + 8

2M - S = -16 + 8

Substitute the value of M,

2(2S/5) - S = -8

-0.2S = -8

S = 40

Now, if we substitute the value of S in the first ratio,

M/S = 2/5

M/40 = 2/5

M = 16

Now, if we take the difference in their ages the difference will be 24(40-16).

Since the difference is 24, we can conclude that Caitlin is right.

Hence, Caitlin is right.

Learn more about Ratios:

brainly.com/question/1504221

#SPJ1

4 0

2 years ago

What is the determinant of the image shown?<br> –14<br> –3<br> 8<br> 11

klio [65]

Answer:-14

Step-by-step explanation:

/A/= (3 x-1)- (1 x 11)

= -3 -11= -14

3 0

3 years ago

Karen received a $80 gift card for a coffee store. She used it in buying some coffee that cost $8.23 per pound. After buying the

aleksandrvk [35]

Answer:

Step-by-step explanation:

80 - 47.08 = 32.92

32.92/8.23 = 4

4 0

3 years ago

Read 2 more answers

If cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2π, what are the values of sin Θ and tan Θ? sin Θ = square root 2 ov

densk [106]

Answer:

$\huge\boxed{\sin\theta=-\dfrac{\sqrt2}{2};\ \tan\theta=-1}$

Step-by-step explanation:

We have:

$\\cos\theta=\dfrac{\sqrt2}{2},\ \dfrac{3\pi}{2}$

For sine use:

$\sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x$

Substitute:

$\sin^2\theta=1-\left(\dfrac{\sqrt2}{2}\right)^2\\\\\sin^2\theta=1-\dfrac{(\sqrt2)^2}{2^2}\\\\\sin^2\theta=1-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4}{4}-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4-2}{4}\\\\\sin^2\theta=\dfrac{2}{4}\to\sin\theta=\pm\sqrt{\dfrac{2}{4}}\\\\\sin\theta=\pm\dfrac{\sqrt2}{\sqrt4}\\\\\sin\theta=\pm\dfrac{\sqrt2}{2}$

θ in IV quadrant, therefore sine is negative.

$\sin\theta=-\dfrac{\sqrt2}{2}$

For tangent use:

$\tan x=\dfrac{\sin x}{\cos x}$

Substitute:

$\tan\theta=\dfrac{-\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}=-\dfrac{\sqrt2}{2}\cdot\dfrac{2}{\sqrt2}=-1$

8 0

3 years ago