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frosja888 [35]
3 years ago
15

Drew jumped farther than all 4 students above but jumped shorter than 7 feet , 7 inches. How far could drew have jumped? Write s

entence explaining how you know that you are correct

Mathematics
1 answer:
S_A_V [24]3 years ago
4 0
<h3>The distance covered by Drew in long jump is either 7 ft 5 inches OR 7 ft  6 inches.</h3>

Step-by-step explanation:

Here, the needed table is attached for the reference.

Now, as we can see from the table:

The distance covered by  Cindy  = 2 yards, 1 foot 3 inches

The distance covered by  Tyrette   =  7 feet, 2 inches

The distance covered by Nina   =   2 yards, 1 foot, 1 inch

The distance covered by  Monique   = 7 feet, 4 inches

As we know : 1 yard  =  3 feet

So, the distance covered  by :

Cindy  =  2 yards, 1 foot 3 inches = 3 ft x (2)  + 1 ft +  3 in  = 7 ft 3 in

Tyrette   =  7 feet, 2 inches

Nina   =   2 yards, 1 foot, 1 inch  = = 3 ft x (2)  + 1 ft +  1 in  = 7 ft 1 in

Monique   = 7 feet, 4 inches

So by comparing all distances, we can see that:

The maximum distances jumped by all four  is 7 ft 4 inches.

The distance covered by Drew is less than 7 ft 7 in.

So, he must have jumped 7 ft 5 inches OR 7 ft  6 inches.

Hence, the distance covered by Drew in long jump is either 7 ft 5 inches OR 7 ft  6 inches.

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Answer:

Amount \ left = 1345.81

Step-by-step explanation:

Total  \  earning = T= $3,511.95

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5 0
3 years ago
-3x^{2}-21x-54 what are the zeros? (Solutions)
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The only way to solve if it is equal to something
assuming that the teacher wanted you to make it equal to zero do
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remember if we can do
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0=-3x^2-21x-54
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remember 0 times anything=0 so
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use quadratice formula which is

if you have
ax^2+bx+c=0 then
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x=\frac{-7+/- \sqrt{7^{2}-4(1)(18)} }{2(1)}
x=\frac{-7+/- \sqrt{49-72} }{2}
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i=√-1
x=\frac{-7+/- i\sqrt{23} }{2}



the zerose would be
x=\frac{-7+ i\sqrt{23} }{2} or \frac{-7- i\sqrt{23} }{2}




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Answer:

Here's your answer

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Answer:

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