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3 years ago

8

Can someone help me with this math problem? I have no idea where to begin. Solve: |y + 8| ≥ 3

1 answer:

lorasvet [3.4K]3 years ago

7 0

Since y+8 is positive, you don't have to find the absolute value since the absolute value is y+8. all you have to do is find a number that makes the statement true for y. the symbol between next to the 3 asks for a number that is greater or equal to 3. So your answer would be

≥-5

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From trigonometry we know that:

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then, $\theta=n\pi+(-1)^n\alpha$ (where $n$ is an integer)

This can be rewritten in degrees as:

$\theta=n(180^{\circ})+(-1)^n\alpha$ .............(Equation 1)

Now, in our case, $\alpha=-230^{\circ}$

Therefore, (Equation 1) can be written as:

$\theta=n(180^{\circ})+(-1)^n(-230^{\circ})$ ..........(Equation 2)

Now, to find the correct options all that we have to do is replace n by relevant integers and find the values of $\theta$ that match.

For n=2, (Equation 2) gives us: $\theta=2\times 180^{\circ}+(-1)^2(-230^{\circ})=360^{\circ}-230^{\circ}=130^{\circ}$ .

Thus, $sin(230^{\circ})=sin(130^{\circ})$

Now, we know that: $-sin(-50^{\circ})=sin(50^{\circ})$

Let n=-1, then:

$\theta=(-1)\times 180^{\circ}+(-1)^{-1}(-230^{\circ})=-180^{\circ}+230^{\circ}=50^{\circ}$

Thus, $sin(-230^{\circ})=-sin(-50^{\circ})$

Likewise, $sin(-230^{\circ})=sin(50^{\circ})$

Only the last option $sin(-50^{\circ})$ will never match $sin(-230^{\circ})$ because no integral value of $n$ will ever give $\theta=-50^{\circ}$

Thus the last option is the correct option.

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