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madam [21]
3 years ago
5

Im having trouble with this one HELP PLZ i want to go to sleep ;-; PLZLZZZZZZZZZ

Mathematics
1 answer:
Yanka [14]3 years ago
7 0
1. (x-7)(x-4)
2. (x-12)(x+7)
use foil to check your work
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Answer:

the answer is 6

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I need to solve the equation please help me
lesantik [10]

Answer:

\displaystyle x=\Big\{\frac{7\pi}{6}+2n\pi, \frac{3\pi}{2}+2n\pi, \frac{11\pi}{6}+2n\pi}\Big\}, n\in\mathbb{Z}

Step-by-step explanation:

We are given the equation:

4\sin^2(x)+6\sin(x)+2=0

First, we can divide everything by 2:

2\sin^2(x)+3\sin(x)+1=0

Notice that we have an equation in quadratic form. Namely, if we make a substitution where u = sin(x), we acquire:

2u^2+3u+1=0

Solve for u. Factor:

(2u+1)(u+1)=0

Zero Product Property:

2u+1=0\text{ or } u+1=0

Solving for both cases:

\displaystyle u=-\frac{1}{2}\text{ or } u=-1

And by substitution:

\displaystyle \sin(x)=-\frac{1}{2}\text{ or } \sin(x)=-1

For the first case, recall that sin(x) is -1/2 for every 7π/6 and every 11π/6. Hence, for the first case, our solutions are:

\displaystyle x=\frac{7\pi}{6}+2n\pi \text{ and } x=\frac{11\pi}{6}+2n\pi, n\in\mathbb{Z}

Where n is an integer.

For the second case, sin(x) is -1 for every 3π/2. Thus:

\displaystyle x=\frac{3\pi}{2}+2n\pi

All together, our solutions are:

\displaystyle x=\Big\{\frac{7\pi}{6}+2n\pi, \frac{3\pi}{2}+2n\pi, \frac{11\pi}{6}+2n\pi}\Big\}, n\in\mathbb{Z}

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1.40

Step-by-step explanation:

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Find the square root of 64a^2/9b^2+4+32a/3b​
insens350 [35]

Answer:

the square root of 64a^2/9b^2+4+32a/3b is

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or

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