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3 years ago

V1 - 2 sin cos 0 = cos 0 - sin

1 answer:

kramer3 years ago

4 0

$\sqrt{1 - 2sin θ \cosθ } = \sqrt{ {sin}^{2} θ + {cos}^{2}θ - 2 sinθcosθ } = \sqrt{ {(cosθ - sinθ) }^{2} } = cosθ - sinθ$

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Ghella [55]

Answer:

$\displaystyle \sf h( - 1) = - \frac{ 2}{3}$

Step-by-step explanation:

we are given a function

$\displaystyle \sf h(x) = \frac{2x - 6}{4 {x}^{2} + 8}$

we would like to simplify it for h(-1)

in order to do so

substitute the value of x

$\displaystyle \sf h( - 1) = \frac{2 \cdot - 1- 6}{4 { (- 1)}^{2} + 8}$

by order of PEMDAS

simplify square:

$\displaystyle \sf h( - 1) = \frac{2 \cdot - 1- 6}{4 { ( 1)}^{} + 8}$

simplify multiplication:

$\displaystyle \sf h( - 1) = \frac{ - 2 - 6}{4 { }^{} + 8}$

simplify addition:

$\displaystyle \sf h( - 1) = \frac{ - 2 - 6}{12}$

simplify subtraction:

$\displaystyle \sf h( - 1) = \frac{ - 8}{12}$

reduce fraction:

$\displaystyle \sf h( - 1) = - \frac{ 2}{3}$

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3 years ago

V1 - 2 sin cos 0 = cos 0 - sin​

V1 - 2 sin cos 0 = cos 0 - sin