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nignag [31]
3 years ago
13

A rectangle with an area of 120 in squared has a length of 8 inches longer then two times its width what is the width of the rec

tangle
Mathematics
1 answer:
Vaselesa [24]3 years ago
6 0

Answer:  \\The\: width\: is \: 6 \:  {in}^{2}  \\ \\  Step\: by\: step\: explanation: \\ Let \:  x  \: be \:  the  \: width  \: of  \: the \:  rectangle,\: x > 0\Rightarrow the \: length:2x + 8 \\ We  \: have  \: an \:  equation:x(2x + 8) = 120 \\ \Leftrightarrow 2 {x}^{2}  + 8x  + 8= 128 \\ \Leftrightarrow  {x}^{2}  + 4x   + 4 = 64 \\ \Leftrightarrow {(x + 2)}^{2}  = 64 \\ \Leftrightarrow x + 2 =  \pm8 \\ \Leftrightarrow x = 6 \:  \vee \: x =  - 10 \\ \Rightarrow x = 6 \: (x > 0)

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solniwko [45]
<h3>Answer: Largest value is a = 9</h3>

===================================================

Work Shown:

b = 5

(2b)^2 = (2*5)^2 = 100

So we want the expression a^2+3b to be less than (2b)^2 = 100

We need to solve a^2 + 3b < 100 which turns into

a^2 + 3b < 100

a^2 + 3(5) < 100

a^2 + 15 < 100

after substituting in b = 5.

------------------

Let's isolate 'a'

a^2 + 15 < 100

a^2 < 100-15

a^2 < 85

a < sqrt(85)

a < 9.2195

'a' is an integer, so we round down to the nearest whole number to get a \le 9

So the greatest integer possible for 'a' is a = 9.

------------------

Check:

plug in a = 9 and b = 5

a^2 + 3b < 100

9^2 + 3(5) < 100

81 + 15 < 100

96 < 100 .... true statement

now try a = 10 and b = 5

a^2 + 3b < 100

10^2 + 3(5) < 100

100 + 15 < 100 ... you can probably already see the issue

115 < 100 ... this is false, so a = 10 doesn't work

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3 years ago
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3 years ago
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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

3+2-(6-(1+1)

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1

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