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hichkok12 [17]
4 years ago
8

QUESTION: 7,8 and 11

Mathematics
1 answer:
V125BC [204]4 years ago
3 0
Hey :)
Number 7:
-->Let's find the least common multiple of 4, 10, and 12 by listing again:
4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40... 60
10: 10, 20, 30, 40, 50, 60
12: 12, 24, 36, 48, 60
-->As you can see, the LCM is d or 60
* Just a random thought but did you accidently write 7 for the other one?

Number 8:
-->By factoring the expression, you would be finding a factor that both numbers share (usually the GCF) and divide the whole expression by that number:
"What is the GCF of 54 and 24?"
"6!"
--> So now that you know the GCF or just factor, you can divide the expression by it:
54a+24b
54a(/6)+24b(/6)
9a+4b
--> Sorry if this gets confusing but now you put the end result in parentheses with what you divided it by on the outside
6(9a+4b)
-->Or in other words, answer choice 'c'!

Number 9:
-->Sorry but I can't answer it yet because you didn't say what his recipt is. I can help you once you tell me what's on the receipt :)
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The volume of the box is:

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The mathematical problem is :

Maximize: V = 4*x^3 - 312*x^2 + 5940*x

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x > 0

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In the maximum, the first derivative of V, dV/dx, is equal to zero

dV/dx = 12*x^2 - 624*x + 5940

From quadratic formula

x = \frac{-b \pm \sqrt{b^2 - 4(a)(c)}}{2(a)}

x = \frac{624 \pm \sqrt{(-624)^2 - 4(12)(5940)}}{2(12)}

x = \frac{624 \pm \sqrt{104256}}{24}

x = \frac{624 \pm \sqrt{2^6*3^2*181}}{24}

x = \frac{624 \pm 8*3*\sqrt{181}}{24}

x_1 = \frac{624 + 24*\sqrt{181}}{24}

x_1 = 26 + \sqrt{181}

x_2 = \frac{624 - 24*\sqrt{181}}{24}

x_2 = 26 - \sqrt{181}

But x_1 > 33, then is not the correct answer.

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