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Orlov [11]
4 years ago
11

Find f(-3) when f(x) = -x²+1​

Mathematics
1 answer:
JulsSmile [24]4 years ago
3 0

Answer:

3x²- 1

Step-by-step explanation:

f(-3) when f(x) = -x²+1

(-x²+1) × -3

3x²- 1

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Square root of 9 I'll mark the 1st​
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Answer:

3

Step-by-step explanation:

4 0
3 years ago
If $95 is put in an account that gets 6% and I add $18 at the end of each year, how much will I have at the end of 11 years?
Temka [501]
6% of $95 = $5.70
multiply $5.70 by 11 years = $62.70 <— interest gained over 11 years
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7 0
3 years ago
Use a half-angle identity to find the exact value
Tatiana [17]

Given:

\cos 15^{\circ}

To find:

The exact value of cos 15°.

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Using half-angle identity:

$\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos (x)}{2}}

$\cos \frac{30^{\circ}}{2}=\sqrt{\frac{1+\cos \left(30^{\circ}\right)}{2}}

Using the trigonometric identity: \cos \left(30^{\circ}\right)=\frac{\sqrt{3}}{2}

            $=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}

Let us first solve the fraction in the numerator.

            $=\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2}}

Using fraction rule: \frac{\frac{a}{b} }{c}=\frac{a}{b \cdot c}

            $=\sqrt{\frac {2+\sqrt{3}}{4}}

Apply radical rule: \sqrt[n]{\frac{a}{b}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}

           $=\frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}}

Using \sqrt{4} =2:

           $=\frac{\sqrt{2+\sqrt{3}}}{2}

$\cos 15^\circ=\frac{\sqrt{2+\sqrt{3}}}{2}

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4 years ago
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The answer would be b
6 0
3 years ago
1000 test scores are approximately normally distributed with a mean of 60 and a standard deviation of 8, if a grade of B is assi
DIA [1.3K]

Answer:

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Step-by-step explanation:

I took the test

8 0
3 years ago
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