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3 years ago

PLEASE HELP I NEED ASAP

Mathematics

1 answer:

Natasha2012 [34]3 years ago

4 0

Answer:

Option A

Step-by-step explanation:

Given inequality:

x - 8 ≥ - 3

Solve it:

x ≥ 8 - 3
x ≥ 5

The graph includes the number line to the right of 5 and the 5 is included, so this point is represented by a full dot.

Correct choice is A

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4.4 is what % of 25

kirza4 [7]

4.4 = .044

0.044x25 = 1.1

ANSWER = 1.1

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3 years ago

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Y=Cosx.arcsinx What is the solution

Alinara [238K]

Step-by-step explanation:

Given: $y = cos(sin^{-1}x))$

As the domain for the inverse sine function is 1 ≤ x ≤ 1 as this is the range for the sine function.

The range for the function is the same as the range for the cosine function, 1 ≤ y ≤ 1

$sin^{2}x +cos^{2}x = 1$

So, using the identity $cos(x) = \pm \sqrt{1-sin^{2}(x) }$

$y = \pm \sqrt{1-sin^{2}(sin^{-1}(x)) }$

As the sin and inverse sin function do cancel each other, so only x² is left.

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$y = \pm \sqrt{1-x^{2}$ ; $-1\leq x\leq 1$

Keywords: trigonometric function

Learn more trigonometric function from brainly.com/question/12551115

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4 years ago

Hello guys i need help

Drupady [299]

Step-by-step explanation:

90*8.25=742 rounds to 675 because its the closest. 85-95=10 divided by 2=5+85=90. 9-7.5=1.5 divided by 2=0.75+7.5=8.25. thats how i got those numbers. THE ANSWER IS C

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3 years ago

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Slope of a line that contains j(1,-4) k(3,-1)

Sauron [17]

M=(y2-y1)/(x2-x1)
=[-1-(-4)] / (3-1)
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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