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2 years ago

Find the EQUATION of the line through (-10,-12) and (-15,3). Use the form y=mx+b.

Mathematics

1 answer:

vovikov84 [41]2 years ago

8 0

Y=Mx+b

Y=-3x-42

Please mark brainliest. Thank you

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2x + 9 = 23 solve by trial

Allisa [31]

Answer:

i tried it by trial and error, and my work is below.

Step-by-step explanation:

2x + 9 = 23

subtract 9 from both sides

2x = 14

guess.... lol

x=7

8 0

3 years ago

There are 26 third graders and 32 fourth graders going on a field trip. each van carry 10 students. how many students are needed

Zolol [24]

26+32=58
6*10=60
6 vans are needed.
10 students will be in the each of the five vans. 8 students will be in the other van.

6 0

3 years ago

Read 2 more answers

Find the solution set of 5x<36 when x ∈ n

Margaret [11]

Answer:

5x<36

5x/5 < 36/5

x < 7.2

4 0

3 years ago

Read 2 more answers

Which of the following is the correct factored form for the given equation? 2x^2+8x+6=0

3241004551 [841]

It would be 2(x+3)(x+1)=0

Explanation:
I used factor by grouping. You multiply the first term (2) by the last term (6). This gives you 12 then take the factors of 12 that add up to the middle term 8. You get 6 and 2.
It should look like 2x^2+6x+2x+6=0
when you do factor by grouping you factor the first two terms and then the last two terms separately. So you get (2x+2) and (x+3). (2x+2) could be factored into 2(x+1). Then you put everything together and get 2(x+3)(x+1)=0

6 0

3 years ago

Find all solutions of each equation on the interval 0 ≤ x < 2π.

Korvikt [17]

Answer:

$x = 0$ or $x = \pi$ .

Step-by-step explanation:

How are tangents and secants related to sines and cosines?

$\displaystyle \tan{x} = \frac{\sin{x}}{\cos{x}}$ .

$\displaystyle \sec{x} = \frac{1}{\cos{x}}$ .

Sticking to either cosine or sine might help simplify the calculation. By the Pythagorean Theorem, $\sin^{2}{x} = 1 - \cos^{2}{x}$ . Therefore, for the square of tangents,

$\displaystyle \tan^{2}{x} = \frac{\sin^{2}{x}}{\cos^{2}{x}} = \frac{1 - \cos^{2}{x}}{\cos^{2}{x}}$ .

This equation will thus become:

$\displaystyle \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} \cdot \frac{1}{\cos^{2}{x}} + \frac{2}{\cos^{2}{x}} - \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} = 2$ .

To simplify the calculations, replace all $\cos^{2}{x}$ with another variable. For example, let $u = \cos^{2}{x}$ . Keep in mind that $0 \le \cos^{2}{x} \le 1 \implies 0 \le u \le 1$ .