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3 years ago

8

Solve the equations. Make conclusion. 3|x+12|=6 Please put down the steps also.

1 answer:

Blababa [14]3 years ago

3 0

3 | x+12 |=6
Start by distributing 3 to both x and 12
You get 3x+36=6
Subtract 36 from both sides
You then get 3x=-30
Divide 3 from both sides getting
x= -10

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A bag has 1 red marble, 4 blue marbles, and 3 green marbles. Peter draws a marble randomly from the bag, replaces it, and then d

Lynna [10]

The total number of marbles in the bag is 1 + 4 + 3 = 8

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3 years ago

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[7.07] Choose the correct product of (6x + 2)2.

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Answer:
36x^2 + 24x + 4

Explanation:
The perfect square has the following general formula:
(a + b)^2 = a^2 + 2ab + b^2

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3 years ago

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7x-6y=9<br> 5x+2y=19 Solve the system using elimination.

ikadub [295]

Step-by-step explanation:

multiply 5x+2y=19 by 3 youll get 15x+6y=57

eliminate y

7x-6y=9

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22x=66

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substitute x to 5x+2y=19

15+2y=19

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3 years ago

Factor the following expression. 3x^3-12x^2-2x-8

den301095 [7]

(x–4) (3x^2 + 2)

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2 years ago

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If 180° < α < 270°, cos⁡ α = −817, 270° < β < 360°, and sin⁡ β = −45, what is cos⁡ (α + β)?

eduard

Answer:

$cos(\alpha+\beta)=-\frac{84}{85}$

Step-by-step explanation:

we know that

$cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)$

Remember the identity

$cos^{2} (x)+sin^2(x)=1$

step 1

Find the value of $sin(\alpha)$

we have that

The angle alpha lie on the III Quadrant

so

The values of sine and cosine are negative

$cos(\alpha)=-\frac{8}{17}$

Find the value of sine

$cos^{2} (\alpha)+sin^2(\alpha)=1$

substitute

$(-\frac{8}{17})^{2}+sin^2(\alpha)=1$

$sin^2(\alpha)=1-\frac{64}{289}$

$sin^2(\alpha)=\frac{225}{289}$

$sin(\alpha)=-\frac{15}{17}$

step 2

Find the value of $cos(\beta)$

we have that

The angle beta lie on the IV Quadrant

so

The value of the cosine is positive and the value of the sine is negative

$sin(\beta)=-\frac{4}{5}$

Find the value of cosine

$cos^{2} (\beta)+sin^2(\beta)=1$

substitute

$(-\frac{4}{5})^{2}+cos^2(\beta)=1$

$cos^2(\beta)=1-\frac{16}{25}$

$cos^2(\beta)=\frac{9}{25}$

$cos(\beta)=\frac{3}{5}$

step 3

Find cos⁡ (α + β)

$cos(\alpha+\beta)=cos(\alpha)*cos(\beta)-sin(\alpha)*sin(\beta)$

we have

$cos(\alpha)=-\frac{8}{17}$

$sin(\alpha)=-\frac{15}{17}$

$sin(\beta)=-\frac{4}{5}$

$cos(\beta)=\frac{3}{5}$

substitute

$cos(\alpha+\beta)=-\frac{8}{17}*\frac{3}{5}-(-\frac{15}{17})*(-\frac{4}{5})$

$cos(\alpha+\beta)=-\frac{24}{85}-\frac{60}{85}$

$cos(\alpha+\beta)=-\frac{84}{85}$

4 0

3 years ago