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scoundrel [369]

3 years ago

14

19. INSECTS An average aunt is 1/4 inch long. An average aphid is 3/32 inch long. How many times longer is an average and then a

n average aphid?
20. LAND A Field has an area of 9/20 Square mile. Find the width of the field if The length is 9/10 miles long.

1 answer:

DaniilM [7]3 years ago

7 0

19 . 2.667 times larger

20. .5

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Lia wants to estimate 4 1/7 x 3 3/5 To estimate, she simplifies the expression 4 x 3 1/2 .

Temka [501]

Given:
4 1/7 * 3 3/5

4 1/7 = (4*7)+1/7 = 29/7
3 3/5 = (3*5)+3/5 = 18/5

29/7 * 18/5 = 29 * 18 / 7 * 5 = 522 / 35 = 14.91

Simplified expression.
4 x 3 1/2

3 1/2 = (3*2)+1/2 = 7/2

4 * 7/2 = 4*7 / 2 = 28 / 2 = 14

It is a close estimate. Actual: 14.91 vs Estimate: 14

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Simplify seven square root of three end root minus four square root of six end root plus square root of forty eight end root min

Cloud [144]

$7\sqrt{3}- 4\sqrt{6} + \sqrt{48} - \sqrt{54}$ simplified as $11\sqrt{3}- 7\sqrt{6}$ or $1.909$ .

<u>Step-by-step explanation:</u>

We need to Simplify seven square root of three end root minus four square root of six end root plus square root of forty eight end root minus square root of fifty four. Which is equivalent to $7\sqrt{3}- 4\sqrt{6} + \sqrt{48} - \sqrt{54}$ :

$7\sqrt{3}- 4\sqrt{6} + \sqrt{48} - \sqrt{54}$

⇒ $7\sqrt{3}- 4\sqrt{6} + \sqrt{48} - \sqrt{44}$

⇒ $7\sqrt{3}- 4\sqrt{6} + \sqrt{16(3)} - \sqrt{9(6)}$

⇒ $7\sqrt{3}- 4\sqrt{6} + 4\sqrt{(3)} - 3\sqrt{(6)}$

⇒ $11\sqrt{3}- 4\sqrt{6}- 3\sqrt{(6)}$

⇒ $11\sqrt{3}- 7\sqrt{6}$

$\sqrt{3} = 1.732 , \sqrt{6} = 2.449$

⇒ $11(1.723)- 7(2.449)$

⇒ $1.909$

Therefore, $7\sqrt{3}- 4\sqrt{6} + \sqrt{48} - \sqrt{54}$ simplified as $11\sqrt{3}- 7\sqrt{6}$ or $1.909$ .

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kari74 [83]

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