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3 years ago

Simplify the expression (1 - 5t) (-8) =

Mathematics

1 answer:

sattari [20]3 years ago

6 0

-8+40t

Step-by-step explanation:

You can solve this by distributing the -8 to both the 1 and -5t. Distributing the -8 to the 1 would get you -8, and distributing it to the -5t would get you 40t, as two negatives multiplied together make a positive.

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Andreas93 [3]

Y = 1.50x + 4......x = 7
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y = 10.50 + 4
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4 years ago

Jenny is covering the top of her freshly made cake with a piece of aluminum foil. The diameter of the cake is 5 inches. What is

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Answer:

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Step-by-step explanation:

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3 years ago

If cos Θ = square root 2 over 2 and 3 pi over 2 < Θ < 2π, what are the values of sin Θ and tan Θ? sin Θ = square root 2 ov

densk [106]

Answer:

$\huge\boxed{\sin\theta=-\dfrac{\sqrt2}{2};\ \tan\theta=-1}$

Step-by-step explanation:

We have:

$\\cos\theta=\dfrac{\sqrt2}{2},\ \dfrac{3\pi}{2}$

For sine use:

$\sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x$

Substitute:

$\sin^2\theta=1-\left(\dfrac{\sqrt2}{2}\right)^2\\\\\sin^2\theta=1-\dfrac{(\sqrt2)^2}{2^2}\\\\\sin^2\theta=1-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4}{4}-\dfrac{2}{4}\\\\\sin^2\theta=\dfrac{4-2}{4}\\\\\sin^2\theta=\dfrac{2}{4}\to\sin\theta=\pm\sqrt{\dfrac{2}{4}}\\\\\sin\theta=\pm\dfrac{\sqrt2}{\sqrt4}\\\\\sin\theta=\pm\dfrac{\sqrt2}{2}$

θ in IV quadrant, therefore sine is negative.

$\sin\theta=-\dfrac{\sqrt2}{2}$

For tangent use:

$\tan x=\dfrac{\sin x}{\cos x}$

Substitute:

$\tan\theta=\dfrac{-\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}=-\dfrac{\sqrt2}{2}\cdot\dfrac{2}{\sqrt2}=-1$

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3 years ago

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Oxana [17]

Answer:

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Step-by-step explanation:

Doing what the problem statement tells you to do, you get ...

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