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3 years ago

11

PLEASE HELP I NEED TO PASS THIS TEST!!! (first actual answer gets brainliest)

1 answer:

Fed [463]3 years ago

8 0

Answer:

W'(4, -2), X'(3, -4), Y'(1, -4), Z'(0, -2)

I hope this helps you. If you can give brainliest that would be greatly appreciated.

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Four pounds of bananas cost $1.96. how much does it cost per pound?

adelina 88 [10]

4 pounds = $1.96
1.96 / 4 = $0.49

4 0

3 years ago

Suppose that 8,000 is placed in an account that pays 17% interest compounded each year assume that no withdraws are made from th

Zarrin [17]

Step-by-step explanation:

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3 0

3 years ago

6/30=3/15 true or false

Dennis_Churaev [7]

Answer: True

Step-by-step explanation: This is because they are equivalent fractions

Hope this helps and have a nice day

4 0

2 years ago

Read 2 more answers

Solve the system of equations.<br> <br><br> 2x−9y=14<br> x=−6y+7<br>

ipn [44]

Step-by-step explanation:

$put \: x \: into \: equation \: 1$

$2( - 6y + 7) - 9y = 14$

$- 12y + 14 - 9y = 14$

$- 21y + 14 = 14$

$- 21y = 14 - 14$

$- 21y = 0$

$\frac{ - 21y}{12} = \frac{0}{21}$

$y = 0$

$put \: y = 0 \: into \: equation \: 2$

$x = 6y + 7$

$x = 6(0) + 7$

$x = 0 + 7$

$x = 7$

3 0

2 years ago

Read 2 more answers

(1+cos2x)/(1-cos2x) = cot^2x

sesenic [268]

We will turn the left side into the right side.

$\dfrac{1 + \cos 2x}{1 - \cos2x} = \cot^2 x$

Use the identity:

$\cos 2x = \cos^2 x - \sin^2 x$

$\dfrac{1 + \cos^2 x - \sin^2 x}{1 - ( \cos^2 x - \sin^2 x)} = \cot^2 x$

$\dfrac{1 - \sin^2 x + \cos^2 x }{1 - \cos^2 x + \sin^2 x} = \cot^2 x$

Now use the identity

$\sin^2 x + \cos^2 x = 1$ solved for sin^2 x and for cos^2 x.

$\dfrac{\cos^2 x + \cos^2 x }{\sin^2 x + \sin^2 x} = \cot^2 x$

$\dfrac{2\cos^2 x}{2\sin^2 x} = \cot^2 x$

$\dfrac{\cos^2 x}{\sin^2 x} = \cot^2 x$

$\cot^2 x = \cot^2 x$

8 0

3 years ago