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Makovka662 [10]
3 years ago
12

Question 2 Please

Mathematics
1 answer:
NeX [460]3 years ago
6 0

Answer:

Step-by-step explanation:

Start by moving the 3 to the other side of the equals sign to get

tan^2(2\theta)=3

tan^2(2\theta)=(tan(2\theta))(tan(2\theta)) so with that replacement:

(tan(2\theta))(tan(2\theta))=3

By the Zero Product Property,

tan(2\theta)=3 or tan(2\theta)=3

Yes these are both the same, so we only need one of them.  Once we find the angle, they are both the same, so we only need one of these equations.

If

tan(2\theta)=3, take the inverse tangent of both sides to give you:

2\theta=tan^{-1}(3) and

2\theta=71.565 so

\theta=35.8

Tangent increases in increments of pi or 180, so the other angle within the given interval is 35.8 + 180 = 215.8

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1/3x +4

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The purpose of the inverse property of multiplication is to get a product of one. So the inverse multiplied with the original number (or equation) should give us the product of one.

1 / (3x + 4) * 3x + 4 = 3x+4 / 3x+ 4 = 1

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Solve the following system
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Answer:

{x = -4 , y = 2 ,  z = 1

Step-by-step explanation:

Solve the following system:

{-2 x + y + 2 z = 12 | (equation 1)

2 x - 4 y + z = -15 | (equation 2)

y + 4 z = 6 | (equation 3)

Add equation 1 to equation 2:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - 3 y + 3 z = -3 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Divide equation 2 by 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+y + 4 z = 6 | (equation 3)

Add equation 2 to equation 3:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+0 y+5 z = 5 | (equation 3)

Divide equation 3 by 5:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y + z = -1 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 3 from equation 2:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x - y+0 z = -2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + y + 2 z = 12 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract equation 2 from equation 1:

{-(2 x) + 0 y+2 z = 10 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Subtract 2 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = 8 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = -4 | (equation 1)

0 x+y+0 z = 2 | (equation 2)

0 x+0 y+z = 1 | (equation 3)

Collect results:

Answer:  {x = -4 , y = 2 ,  z = 1

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Answer:

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