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Sliva [168]
3 years ago
5

Solve for x 2cos^2x-3cosx+1=0

Mathematics
1 answer:
Romashka [77]3 years ago
4 0
<span> x= Pi, x = 2pi/3, x = 4pi/3
That is the answer</span>
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Expand the following 2(x+3)
Maksim231197 [3]

Answer:

2x + 6

Step-by-step explanation:

2 will multiply everything inside the bracket.

2 x X =2x

2 X 3=6

6 0
2 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=%5Clarge%5Cmathfrak%7B%7B%5Cpmb%7B%5Cunderline%7B%5Corange%7BQuestion%20%7D%7D%7B%5Corange%7B%
VashaNatasha [74]

Answer:

29

Step-by-step explanation:

As it is a list of prime number, the sequence continues with 29 being the next prime number after 23.

6 0
3 years ago
I need a long essay about may the bridges i burn light the way​
aivan3 [116]

Answer:

I will let it lead me down the paths ahead of me. I never in my life thought that Dylan McKay would be a major influence on my growth as a person. But that's the way it seems to be going right now. So wish me luck, and may the bridges I've burned light the way.

this is not that long but I tried

6 0
2 years ago
HJ
Olin [163]

Answer:

19 degrees

Step-by-step explanation:

From the question given

The interior angles are x+12 and x - 3

Exterior angle is <IJK = 5x-6

Using the rule that states that the sum of interior angle of a triangle is equal to the exterior

<JHI + <HIJ = <IJK

x+12 + x-3 = 5x - 6

2x+9 = 5x -6

2x - 5x = -6-9

-3x = -15

x = -15/-3

x = 5

Get <IJK

Recall that <IJK = 5x - 6

<IJK = 5(5) - 6

<IJK = 25-6

<IJK = 19 degrees

Hence the measure of <IJK is 19 degrees

8 0
3 years ago
(1.)Find the slope of the line that passes through the given pair of points. (If an answer is undefined, enter UNDEFINED.) (?a +
jekas [21]

Answer:

1. The slope of the line is m=\frac{-2b+3}{2a}.

2. The value of a is 18.

Step-by-step explanation:

If a line passes through two points, then the slope of the line is

m=\frac{y_2-y_1}{x_2-x_1}

(1)

It is given that the line passes through the points (-a + 3, b - 3) and (a + 3, -b). So, the slope of the line is

m=\frac{-b-(b-3)}{a+3-(-a+3)}

m=\frac{-b-b+3}{a+3+a-3)}

m=\frac{-2b+3}{2a}

The slope of the line is m=\frac{-2b+3}{2a}.

(2)

If the line passing through the points (a, 1) and (6, 5), then the slope of the line is

m_1=\frac{5-1}{6-a}=\frac{4}{6-a}

If the line passing through the points (2, 7) and (a + 2, 1), then the slope of the line is

m_2=\frac{1-7}{a+2-2}=\frac{-6}{a}

The slopes of two parallel lines are same.

m_1=m_2

\frac{4}{6-a}=\frac{-6}{a}

On cross multiplication we get

4a=-6(6-a)

4a=-36+6a

4a-6a=-36

-2a=-36

Divide both sides by -2.

a=18

Therefore the value of a is 18.

8 0
3 years ago
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