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3 years ago

Anyone know how to solve this??

Mathematics

1 answer:

Tom [10]3 years ago

7 0

$Given : \frac{(x^\frac{2}{5})^9\;.\;(x^\frac{-4}{15})}{x^\frac{1}{3}}$

$\implies {(x^\frac{18}{5})\;.\; (x^\frac{-4}{15})}\;.\;(x^\frac{-1}{3})$

$\implies x^(^\frac{18}{5} ^-^\frac{4}{15}^-^\frac{1}{3}^)$

$\implies x^(^(^\frac{54 - 4}{15}^)^-^\frac{1}{3}^) = x^(^(^\frac{50}{15}^)^-^\frac{1}{3}^) = x^(^(^\frac{10}{3}^)^-^\frac{1}{3}^) = x^(^\frac{9}{3}^) = x^3$

$\implies x^k = x^3$

$\implies k = 3$

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2 years ago

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3 years ago

6. Two observers, 7220 feet apart, observe a balloonist flying overhead between them. Their measures of the

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Answer:

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Step-by-step explanation:

Let's call:

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$a_2$ the horizontal distance between the second observer and the ballonist

$\alpha _1$ and $\alpha _2$ the angles of elevation meassured by each observer

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So, the area of a triangle is the length of its base times its height.

$S=a*h$ (equation 1)

but we can divide the triangle in two right triangles using the height line. So the total area will be equal to the addition of each individual area.

$S=S_1+S_2$ (equation 2)

$S_1=a_1*h$

But we can write $S_1$ in terms of $\alpha _1$ , like this:

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And for $S_2$ will be the same:

$S_2=\frac{h^{2} }{\tan(\alpha _2)}$

Replacing in the equation 2:

$S=\frac{h^{2} }{\tan(\alpha _1)}+\frac{h^{2} }{\tan(\alpha _2)}\\S=h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})$

And replacing in the equation 1:

$h^{2}*(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})=a*h\\h=\frac{a}{(\frac{1 }{\tan(\alpha _1)}+\frac{1}{\tan(\alpha _2)})}$

So, we can replace all the known data in the last equation:

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Then, the ballonist is at a height of 3579.91 ft above the ground at 3:30pm.

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