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3 years ago

9

For a given input value x, the function h outputs a value y to satisfy the following equation 6x+y=4x+11y. Write a formula for h

(x) in the terms of x

1 answer:

Tanzania [10]3 years ago

5 0

Answer: h(x) = $\frac{1}{5}x$

<u>Step-by-step explanation:</u>

6x + y = 4x + 11y

<u> -y </u> <u> -y </u>

6x = 4x + 10y

<u>-4x </u> <u>-4x </u>

2x = 10y

<u>÷10 </u> <u> ÷10 </u>

$\frac{1}{5}x$ = y

Since "y" represents h(x), then h(x) = $\frac{1}{5}x$

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2 years ago

60 + 84<br> the sum of the numbers as a product of their GCF is?

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Find the common ratio and the next term for the sequence shown. Express your answers as fractions or as decimals.

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3 years ago

Find cos(2*ABC) 100POINTS

juin [17]

Answer:

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Step-by-step explanation:

<u>Trigonometric Identities</u>

$\cos(A \pm B)=\cos A \cos B \mp \sin A \sin B$

<u>Trigonometric ratios</u>

$\sf \sin(\theta)=\dfrac{O}{H}\quad\cos(\theta)=\dfrac{A}{H}\quad\tan(\theta)=\dfrac{O}{A}$

where:

$\theta$ is the angle
O is the side opposite the angle
A is the side adjacent the angle
H is the hypotenuse (the side opposite the right angle)

Using the trig ratio formulas for cosine and sine:

$\cos(\angle ABC)=\dfrac{3}{5}$

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Therefore, using the trig identities and ratios:

$\begin{aligned}\implies \cos(2 \cdot \angle ABC) & = \cos(\angle ABC + \angle ABC)\\\\& = \cos (\angle ABC) \cos (\angle ABC) - \sin(\angle ABC) \sin (\angle ABC)\\\\& = \cos^2(\angle ABC)-\sin^2(\angle ABC)\\\\& = \left(\dfrac{3}{5}\right)^2-\left(\dfrac{4}{5}\right)^2\\\\& = \dfrac{3^2}{5^2}-\dfrac{4^2}{5^2}\\\\& = \dfrac{9}{25}-\dfrac{16}{25}\\\\& = \dfrac{9-16}{25}\\\\& = -\dfrac{7}{25} \end{aligned}$

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