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2 years ago

Can someone help me solve this problem? (photo)

Mathematics

2 answers:

Stolb23 [73]2 years ago

7 0

Answer:

$- \sqrt{ - (2 \times 2 \times 2 \times 2 \times 3)}$

$4 \sqrt{ - 3}$

hope it's helps

answer is B)-4root3

pishuonlain [190]2 years ago

3 0

Answer:

$-4i\sqrt{3}$

Step-by-step explanation:

$\sqrt{-1} =i$

$\sqrt{a} \sqrt{b} =\sqrt{ab}$

$-\sqrt{-48} =-\sqrt{-1} \sqrt{48} =-i\sqrt{16} \sqrt{3} =-4i\sqrt{3}$

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3 years ago

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DanielleElmas [232]

Answer:

Step-by-step explanation:

First, convert all the secants and cosecants to cosine and sine, respectively. Recall that $csc(x)=1/sin(x)$ and $sec(x)=1/cos(x)$ .

Thus:

$\frac{sec(x)}{cos(x)} -\frac{sin(x)}{csc(x)cos^2(x)}$

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Let's do the first part first: (Recall how to divide fractions)

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For the second term:

$\frac{sin(x)}{\frac{cos^2(x)}{sin(x)} } =\frac{sin(x)}{1} \cdot\frac{sin(x)}{cos^2(x)}=\frac{sin^2(x)}{cos^2(x)}$

So, all together: (same denominator; combine terms)

$\frac{1}{cos^2(x)}-\frac{sin^2(x)}{cos^2(x)}=\frac{1-sin^2(x)}{cos^2(x)}$

Note the numerator; it can be derived from the Pythagorean Identity:

$sin^2(x)+cos^2(x)=1; cos^2(x)=1-sin^2(x)$

Thus, we can substitute the numerator:

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Everything simplifies to 1.

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3 years ago

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2 years ago

Can someone help me solve this problem? (photo) ​

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