All categories

3 years ago

What does W prime V prime and U prime equal

Mathematics

1 answer:

choli [55]3 years ago

5 0

Answer:

W' = (0, 1.5)

V' = (1, 2)

U' = (1.5, 0)

Step-by-step explanation:

Multiply each coordinate of the given points by the scale factor 0.5

0*0.5 = 0

3*0.5 = 1.5

2*0.5 = 1

4*0.5 = 2

So thats why W(0,3) becomes W'(0, 1.5), and why V(2,4) becomes V'(1,2), and why U(3,0) becomes U'(1.5, 0)

You might be interested in

James took a total of 28 quizzes over the course of 4 weeks. How many

Andru [333]

Answer:

Step-by-step explanation:

We use rates to solve this so we do 28/4=7.

Using that, we know he can do 7 quizzes in 1 week.

Now 7 times 6= 42 right? so do 1 times 6.

The final answer is 6 weeks.

Hope this helps :)

3 0

2 years ago

Subtract<br><br> 4 hundreds 6 hundreths - 3 hundreths = __________ hundreds __________ hundreths

inn [45]

The answer would be 4 hundreds 3 hundreths

6 0

2 years ago

Read 2 more answers

Simplify.<br> 52 • (4 + 23)

borishaifa [10]

Answer:

Step-by-step explanation:

To use the order of operations stands for (Parenthesis, Exponents, Multiply, Divide, Add, and Subtract) left to right. Please Excuse My Dear Aunt Sally!

BOMDAS stands for (Brackets, Of, Multiply, Divide, Add, and Subtract) from left to right.

First, do parenthesis.

52*(4+23)

4+23=27

52*27

Then, multiply.

52*27=1404

1404

The correct answer is 1404.

4 0

3 years ago

A company pays employees 55 cents per mile to drive their own cars for business trips. If you drive 176 miles on a business trip

makkiz [27]

Answer:

96.80

Step-by-step explanation:

All you do is 176 by 55

3 0

2 years ago

Read 2 more answers

Factor completely: <br><br> A) 64x²-49y²<br> B) 20x²-45y²

tigry1 [53]

$Use\\a^2-b^2=(a-b)(a+b)\\(ab)^n=a^nb^n\\(\sqrt{a})^2=a\ for\ a\geq0\\------------------------------\\A)\ 64x^2-49y^2=8^2x^2-7^2y^2=(8x)^2-(7y)^2=(8x-7y)(8x+7y)\\\\B)\ 20x^2-45y^2=(\sqrt{20})^2x^2-(\sqrt{45})^2y^2=(*)\\\\\sqrt{20}=\sqrt{4\cdot5}=\sqrt4\cdot\sqrt5=2\sqrt5\\\sqrt{45}=\sqrt{9\cdot5}=\sqrt9\cdot\sqrt5=3\sqrt5\\\\(*)=(2\sqrt5x)^2-(3\sqrt5y)^2=(2\sqrt5x-3\sqrt5y)(3\sqrt5x+3\sqrt5y)$

$or:\\\\20x^2-45y^2=5\cdot4x^2-5\cdot9y^2=5(4x^2-9y^2)=5[(2x)^2-(3y)^2]\\\\=5(2x-3y)(2x+3y)$

3 0

2 years ago

Read 2 more answers