A lighthouse is 15m above sea level.

1 answer:

8 0

For this you will use the tangent function. You have the adjacent side and have to find the opposite side.

tan x = opposite/adjacent
tan 25 = x/15
15 (tan 25) = x
6.9946 = x

So the boat is approximately 6.9946 m away from the lighthouse.

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The product is $104 x^{4}+16 \sqrt{30} x^{4}$

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The given expression is $\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}$

We need to determine the product of the given expression.

First, we shall simplify the given expression.

Thus, we have,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x \sqrt{5} x+2 x^{2} \sqrt{6}\right)^2$

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)^2$

Expanding the expression, we have,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)$

Now, we shall apply FOIL, we get,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}\right)^{2}+2 ( 2 x^{2} \sqrt{6})(4 x^{2} \sqrt{5})+\left(2 x^{2} \sqrt{6}\right)^{2}$