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3 years ago

(02.02 LC)

Mathematics

2 answers:

Katyanochek1 [597]3 years ago

7 0

The figure was simply rotated, the relative side lengths have not changed.

Length AB=Length A'B'

Elza [17]3 years ago

5 0

Length of AB=Length of A'B' because they are corresponding sides.

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Draw a shape to show 5/2

IRISSAK [1]

This is a tricky question lol. 5/2 = 2.5 , which is half of 5 ... the picture will illustrate half of 5

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3 years ago

What is <br> 8v+w+7−8v+2w

Bingel [31]

Answer:

3w+7

Step-by-step explanation:

Let's simplify step-by-step.

8v+w+7−8v+2w

=8v+w+7+−8v+2w

Combine Like Terms:

=8v+w+7+−8v+2w

=(8v+−8v)+(w+2w)+(7)

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2 years ago

Change the following decimal numbers to percents. .25 .278 2.9 3.0 0.67

umka2103 [35]

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2 years ago

Read 2 more answers

Find the equation of this line.<br> Acellus

aalyn [17]

Answer:

Step-by-step explanation:

two points on the line are (-3,-2) and (5,4)

slope=(4+2)/(5+3)=6/8=3/4

eq. of line is

y+2=3/4 (x+3)

4y+8=3x+9

4y=3x+1

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2 years ago

What is the exact value of arccos(−2√2) ?

Strike441 [17]

Answer:

See explanation

Step-by-step explanation:

The range of the cosine function is $-1\le \cos x\le 1$

Therefore $arc \cos (-2\sqrt{2})$ is not defined.

Assuming your question is rather $arc \cos (-\frac{\sqrt{2}}{2})$

Then $arc \cos (-\frac{\sqrt{2}}{2})=\pi-arc \cos (\frac{\sqrt{2}}{2})$ in the second quadrant.

$arc \cos (-\frac{\sqrt{2}}{2})=\pi-\frac{\pi}{4}$

$arc \cos (-\frac{\sqrt{2}}{2})=\frac{3\pi}{4}$

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3 years ago