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3 years ago

12

How do I find range and domain in algebra (graph)

1 answer:

Ierofanga [76]3 years ago

4 0

Answer:

You can find the range by seeing the y-value of the plotted points.

You can find domain by looking at the x-values of the plotted points.

Hope you found this helpful!!!

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If the volume of the pyramid shown is 12cm^3, what is its height?

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I think it’s either C or D

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Can someone plss help mee ??

Dvinal [7]

First - 8
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3 years ago

Please help anyone and explain please

umka2103 [35]

So if you distribute 4 in the first parentheses, you get 12x+20y+8z.

Then you distribute 3 in the second parentheses. You'll get 3x-3z. That all equals 12x+20y+8z+3x-3z.

Now you have to start combining numbers with the same variable. Start with x. 12x+3x is 15x.

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8z-3z is 5z

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Find the equation of a line that contains the points (4, -2) and (-8, -1). Write the equation in slope-intercept form, using

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y =−12 /1 x−3 /5

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3 years ago

4.) What is the exact value of sinθ when θ lies in Quadrant II and cosθ=−513

dimaraw [331]

Answer:

Part 4) $sin(\theta)=\frac{12}{13}$

Part 10) The angle of elevation is $40.36\°$

Part 11) The angle of depression is $78.61\°$

Part 12) $arcsin(0.5)=30\°$ or $arcsin(0.5)=150\°$

Part 13) $-45\°$ or $225\°$

Step-by-step explanation:

Part 4) we have that

$cos(\theta)=-\frac{5}{13}$

The angle theta lies in Quadrant II

so

The sine of angle theta is positive

Remember that

$sin^{2}(\theta)+ cos^{2}(\theta)=1$

substitute the given value

$sin^{2}(\theta)+(-\frac{5}{13})^{2}=1$

$sin^{2}(\theta)+(\frac{25}{169})=1$

$sin^{2}(\theta)=1-(\frac{25}{169})$

$sin^{2}(\theta)=(\frac{144}{169})$

$sin(\theta)=\frac{12}{13}$

Part 10)

Let

$\theta$ ----> angle of elevation

we know that

$tan(\theta)=\frac{85}{100}$ ----> opposite side angle theta divided by adjacent side angle theta

$\theta=arctan(\frac{85}{100})=40.36\°$

Part 11)

Let

$\theta$ ----> angle of depression

we know that

$sin(\theta)=\frac{5,389-2,405}{3,044}$ ----> opposite side angle theta divided by hypotenuse

$sin(\theta)=\frac{2,984}{3,044}$

$\theta=arcsin(\frac{2,984}{3,044})=78.61\°$

Part 12) What is the exact value of arcsin(0.5)?

Remember that

$sin(30\°)=0.5$

therefore

$arcsin(0.5)$ -----> has two solutions

$arcsin(0.5)=30\°$ ----> I Quadrant

or

$arcsin(0.5)=180\°-30\°=150\°$ ----> II Quadrant

Part 13) What is the exact value of $arcsin(-\frac{\sqrt{2}}{2})$

The sine is negative

so

The angle lies in Quadrant III or Quadrant IV

Remember that

$sin(45\°)=\frac{\sqrt{2}}{2}$

therefore

$arcsin(-\frac{\sqrt{2}}{2})$ ----> has two solutions

$arcsin(-\frac{\sqrt{2}}{2})=-45\°$ ----> IV Quadrant

or

$arcsin(-\frac{\sqrt{2}}{2})=180\°+45\°=225\°$ ----> III Quadrant

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3 years ago