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2 years ago

13

The volume of the shipping box needs to be 1,144 cubic inches. The equation that models the volume of the shipping box is 8(n +

2)(n + 4) = 1,144.
Answer the following questions about the equation modeling the volume of the shipping box.

Question 1

Solve the equation that models the volume of the shipping box, 8(n + 2)(n + 4) = 1,144. If you get two solutions, are they both reasonable?

2 answers:

Snezhnost [94]2 years ago

7 0

Answer:

see below

Step-by-step explanation:

8(n + 2)(n + 4) = 1,144

FOIL

8(n^2 +2n+4n+8) = 1144

Divide each side by 8

8/8(n^2 +2n+4n+8) = 1144/8

(n^2 +2n+4n+8) = 143

Combine like terms

n^2 +6n+8 = 143

Subtract 143 from each side

n^2 +6n+8 -143= 0

Combine like terms

n^2 +6n -135 =0

Factor

What two terms multiply to -135 and add to 6

-9*15 =-135

-9+15 = 6

(n-9) (n+15) =0

Using the zero product property

n-9 =0 n+15=0

n = 9 n=-15

The length cannot be negative so n = -15 cannot be a solution

n =9

g100num [7]2 years ago

7 0

Answer:

n = 9

Step-by-step explanation:

8(n + 2)(n + 4) = 1,144

(n + 2)(n + 4) = 143

n² + 2n + 4n + 8 = 143

n² + 6n - 135 = 0

n² + 15n - 9n - 135 = 0

n(n + 15) - 9(n + 15) = 0

(n - 9)(n + 15) = 0

n = 9, -15

since n is a length, it can not be negative

Therefore the only solution is n = 9

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QUESTION 1

GIVEN:

$long \: base = 3 \times (short \: base)$

$height = (short \: base) - 2$

$area = 30 {in}^{2}$

IMPORTANT EQUATIONS

$trapezoid \: area = \frac{sum \: of \: bases}{2} \times height$

SOLVE:

$30 = \frac{short + long}{2} (short - 2)$
$30 = \frac{(short + 3short)}{2} \times (short - 2)$

$30 = \frac{4short}{2} (short - 2)$

$30 = 2{(short)}^{2} - 4(short)$

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factor:

$(short - 5)(short + 3) = 0$

$short = 3 \: and \: - 5$
Negative five is not a reasonable answer, so we focus on the positive three and say that is the length of the short base.

so, the answers:

$short \: base = 3 \: in$

$long \: base = 3 \times 3 = 9 \: in$

$height = 3 - 2 = 1 \: in$

QUESTION 2

GIVEN
convert 3ft to inches.

$l = 32 + w$

$area = lw = 130 {in}^{2}$
or

$w = \frac{130}{l}$

substitute

$l = 32 + \frac{130}{l}$

solve for l:

${l}^{2} - 32 l- 130 = 0$
QUESTION 3

$x(x + 2) = 80$

solve for X

${x}^{2} + 2x - 80 = 0$
$(x - 10)(x + 8) = 0$

$x = 10 \: and \: - 8$

take the positive value since it's the only one that makes sense in this context.

so the two values are:

$10 \: and \: 8$

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