Help please come asp

Mathematics

1 answer:

Alex787 [66]3 years ago

5 0

6x-18 is the answer.

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Express cos^2 x in terms of cos2x. Hence, express f(x) = 6cos^2 x - 4sin 2x in the form R cos (2x + Y degrees) + p where R, y an

Lady bird [3.3K]

2cos^2 x = 1 + cos 2x
cos^2 x = 1/2(1 + cos 2x)
f(x) = 6cos^2 x - 4 sin 2x = 6(1/2(1 + cos 2x) - 4 sin 2x = 3 + 3cos 2x - 4sin 2x

8 0

3 years ago

How to find variables of these #2 & #3

Lubov Fominskaja [6]

So... hmmm if you check the first picture below, for 2)

we could use the proportions of those small, medium and large similar triangles like $\bf \cfrac{small}{large}\qquad \cfrac{x}{12}=\cfrac{6}{x}\impliedby \textit{solve for "x"}
\\\\\\
\cfrac{small}{large}\qquad \cfrac{z}{18}=\cfrac{6}{z}\impliedby \textit{solve for "z"}
\\\\\\
\cfrac{large}{medium}\qquad \cfrac{y}{12}=\cfrac{18}{y}\impliedby \textit{solve for "y"}$

now.. for 3) will be the second picture below

$\bf \cfrac{large}{medium}\qquad \cfrac{x+10}{2\sqrt{30}}=\cfrac{2\sqrt{30}}{10}\impliedby \textit{solve for "x"}
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\textit{now, because you already know what "x" is, we can use it below}
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\cfrac{large}{small}\qquad \cfrac{z}{x}=\cfrac{x+10}{z}\impliedby \textit{solve for "z"}
\\\\\\
\textit{and let us use "x" again below}
\\\\\\
\cfrac{small}{medium}\qquad \cfrac{y}{10}=\cfrac{x}{y}\impliedby \textit{solve for "y"}$