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loris [4]
2 years ago
10

clare wants to make an open-top box by cutting out corners of a 30 inch by 25 inch piece of poster board and then folding up the

sides. The volume V(x) in cubic inches of the open-top is a function of the side lengh x in inches of the square cutouts. Write an expression for V(x).
Mathematics
1 answer:
OverLord2011 [107]2 years ago
4 0

Answer:

A. V(x)= (50-2x) ( 35-2x)x

Step-by-step explanation:

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If sin A = 3/8, find the value of cosec A - sec A.​
alexira [117]

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\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}

Step by step explanation:

\text{Given that,}\\\\~~~~~~\sin A = \dfrac 38 \\\\\implies \sin^2 A = \dfrac 9{64}\\\\\implies  1 - \cos^2 A = \dfrac{9}{64}\\\\\implies \cos ^2 A = 1 - \dfrac 9{64}\\\\\implies \cos^2 A = \dfrac{55}{64}\\\\\implies \cos A =\pm\sqrt{\dfrac{55}{64}}\\ \\\implies \cos A = \pm\dfrac{\sqrt{55}}8\\\\

\implies \dfrac 1{\cos A} = \pm\dfrac{8}{\sqrt{55}}

\text{Now,}\\\\\csc A - \sec A\\\\=\dfrac{1}{\sin A}- \dfrac{1}{\cos A}\\\\=\dfrac 83 -\left(\pm \dfrac 8{\sqrt {55}} \right)\\ \\\text{Hence,}\\\\\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}

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