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Flura [38]
3 years ago
10

the perimeter of a pool table is 30 ft.The table is twice as long as it is wide.what is the length of the pool table.

Mathematics
1 answer:
noname [10]3 years ago
4 0
Perimeter is legnth +legnth + width + width or
P=2L+2W

L=2W
subsitute L=2W for L in P=2L+2W

P=2(2W)+2W
P=4W+2W
P=6W

we know that Perimiter=30 so
30=6W
divide both sides by 6
5=Width

subsitute W=5 for W in L=2W
L=2(5)
L=10

to check
10=2(10)+2(5)
30=20+10
30=30
checks


Legnght=10
Width=5
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Using the Pythagorean theorem a^2 + b^2 = c^2
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36 + b^2 = 64

b^2 = 64 - 36
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b = sqrt(28)
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b = 5.3 yards



4 0
3 years ago
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Answer:

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How do u slove this and whats the answer. 2b + 9x + (2 + 5)
e-lub [12.9K]

2b + 9x +(2 + 5).  

2b + 9x + 7=2b + 9x + 7

4 0
3 years ago
2 years ago, father age was nine times the son age but 3 years later it will be 5 times only. find the present ages of the fathe
Goshia [24]

Answer:

<u><em>The Father is currently 47 and the Son is 7</em></u>

Step-by-step explanation:

Let F and S be the present ages of Father and Son, respectively.

We are told that <u>(F-2) = 9(S-2)</u> [2 years ago, father age was nine times the son age]

We also learn that <u>(F+3) = 5(S+3)</u>  [3 years later it will be 5 times only]

Take the first expression and isolate one of the variables (S or F).  I'll isolate F:

(F-2) = 9(S-2)

F = 9S - 16

Now use this in the second expression:

(F+3) = 5(S+3)

((9S-16)+3) = 5(S+3)

9S-13 = 5S+15

4S = 28

S = 7

Since F = 9S-16,

F = 9*(7)-16

F = 47

<u><em>Father is 47 and Son is 7</em></u>

CHECK:

Was the father 9 times the age of his son 2 years ago?

Father would have been 45 and son 5.  Yes, 9*5 = 45

In 3 years will he be 5 times older than his son?  Yes, Father would be 50 and son would be 10.  5*(10) = 50

3 0
1 year ago
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