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lina2011 [118]
3 years ago
12

If the width and height of a rectangular prism are each shrunk to one seventh of the original size but the length remains the sa

me, what is the formula to find the modified surface area?
Mathematics
1 answer:
Gekata [30.6K]3 years ago
5 0
Hey there,

Let's say    'L'=length,  'W'=width, and  'H'=height.

<span>The surface area of a rectangular prism is  </span>

                         A  =  (2LW) + (2WH) + (2LH)  =  2 (LW + WH + LH)

<span>If you shrink the width and height, then the new width is  W/7 </span>
and the new height is H/7.

The new surface area is

                         A = (2 L W/7) + (2 W/7 H/7) + (2 L H/7) .

<span>That's               </span>A = 2 (LW/7 + WH/49 + LH/7)<span> .</span>

<span>or you could write it as      </span><span>A = 2/7 (LW + WH/7 + LH)

</span><span>There are many more ways to massage the formula and write it </span>
<span>in different ways, but I think writing it just that way is OK.</span>
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Solve 11 ≤ w + 3.4 i dont know the answer please help me
shepuryov [24]

Answer:

7.6 ≤w

Step-by-step explanation:

Hey there!

In order to solve this inequality, you need to simplify the inequality like the following:

Subtract 3.4 from both sides

7.6 ≤w

This means that w is greater than or equal to 7.6

6 0
2 years ago
Given: PSTK is a trapezoid, m∠P=90° SK =13, PK = 12, ST=8 Find: The area of PSTK.
guajiro [1.7K]
<h3>Given</h3>

trapezoid PSTK with ∠P=90°, KS = 13, KP = 12, ST = 8

<h3>Find</h3>

the area of PSTK

<h3>Solution</h3>

It helps to draw a diagram.

∆ KPS is a right triangle with hypotenuse 13 and leg 12. Then the other leg (PS) is given by the Pythagorean theorem as

... KS² = PS² + KP²

... 13² = PS² + 12²

... PS = √(169 -144) = 5

This is the height of the trapezoid, which has bases 12 and 8. Then the area of the trapezoid is

... A = (1/2)(b1 +b2)h

... A = (1/2)(12 +8)·5

... A = 50

The area of trapezoid PSTK is 50 square units.

5 0
3 years ago
2) Juan had $622 in his checking account. He deposited a check for $180, and $100
adoni [48]

Answer:

867

Step-by-step explanation:

622+180+100=902

902-35=867

3 0
3 years ago
In circle K shown below, points B, C, D, and E lie on the circle with secants HBD and HCE drawn. Prove:
Alexxx [7]

Answer:

By exterior angle theorem, we have;

∠DBE = ∠H + ∠HEB = ∠ECD = ∠H + ∠HDC

∴ ∠H + ∠HEB = ∠H + ∠HDC

By addition property of equality, we have

∠HEB = ∠HDC

∠H = ∠H by reflexive property

∴ ΔHCD ~ ΔHEB by Angle Angle AA similarity postulate

∴ HE/HD = EB/DC, by the definition of similarity

Therefore, by cross multiplication, we have;

HE × DC = EB × HD

Therefore, by commutative property of multiplication, we have;

HE × DC = HD × EB

Step-by-step explanation:

3 0
3 years ago
I need help with any of these please try to explain
Nana76 [90]
Its math just use a calculator or something
6 0
3 years ago
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