Ask question

Ask question

All categories

3 years ago

12

If the cube is divided into two equal parts by a plane parallel to the face defined by vertices 2, 3, 6, and 7, what will be the

area of the cross-section?

1 answer:

notsponge [240]3 years ago

5 0

The area of the cross-section is 169 sq. cm.

That is the correct answer to this problem

You might be interested in

I know it's late but im having trouble in math.I need an exper to help for BRAINLEST and extra points

frez [133]

Answer:

B'(- 2, 2 )

Step-by-step explanation:

Given the translation rule (x, y ) → (x + 3, y - 2 )

This means add 3 to the original x- coordinate and subtract 2 from the original y- coordinate, that is

B(- 5, 4 ) → B'(- 5 + 3, 4 - 2 ) → B'(- 2, 2 )

8 0

2 years ago

There are two x intercepts for y = (x + 2)(x - 1)<br> The first is<br> and the 2nd is

Vsevolod [243]

Answer:

(-2,0) and (1,0)

Hope this helps!

5 0

3 years ago

What is the answer?please help

White raven [17]

Answer:

d

Step-by-step explanation:

4 0

2 years ago

What is the rate for 4 shirts for $32:00

EastWind [94]

The answer should be $8 if you are trying to find the unit rate. All you have to do is divide 32 by 4 and you get 8. Hope that helped.

3 0

3 years ago

Read 2 more answers

the circle below has center T. suppose that m UV = 138 degrees and that UW is tangent to the circle at U. find the following

alisha [4.7K]

Angle UTV is basically given: it's 138°.

Moreover, since UTV is isosceles, the other two angles have the same measure x. We thus have

$138+x+x=180 \iff 2x=42 \iff x=21$

So, TUV is 21°, and VUW is complementary, so it must be 69° so that they sum up to 90°

3 0

3 years ago