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natita [175]
3 years ago
12

For khan academy . Need answer immediately

Mathematics
1 answer:
kolbaska11 [484]3 years ago
8 0

It's easy just use y=mx+b form ok?

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Geometric mean between 9 and 36
nlexa [21]
I've actually just learned this. We would set up the proportion as X/9=36/X so after we cross multiply we get x^2=324. Then we find the square root of both sides to simplify. And using my calculator the square root of x^2 is just x. And the square root of 324 is 18. So the final answer is x=18 or the geometric mean is 18.
4 0
3 years ago
Re-posting this because nobody noticed. Please somebody help me with this. Its my first time on brainly.
Svetlanka [38]

Answer:

1. Should stay the same since you can't combine any further

2. 6x²-6x+16

8 0
3 years ago
Which of the following values of x makes this equation true? (-4x – 3) - (7x - 3) = 22​
Katena32 [7]

Answer:

Step-by-step explanation:

Here you go mate

Step 1

(-4x-3)-(7x-3) =22​  Equation

Step 2

(-4x-3)-(7x-3) =22​  Simplify

-11x=22

Step 3

-11x=22  Divide

Answer

x=-2

5 0
3 years ago
Help me please .............................................
scoundrel [369]

Answer:

what the picture?

Step-by-step explanation:

3 0
3 years ago
How to solve this ? I am not sure pls if you know the answer answer it I really neeed it for marks
xz_007 [3.2K]

Answer:

L.S = R.S ⇒ Proved down

Step-by-step explanation:

Let us revise some rules in trigonometry

  1. sin²α + cos²α = 1
  2. sin2α = 2 sin α cosα
  3. cscα = 1/sinα

To solve the question let us find the simplest form of the right side and the left side, then show that they are equal

∵ L.S = csc2α + 1

→ By using the 3rd rule above

∴ L.S = \frac{1}{sin2\alpha} + 1

→ Change 1 to \frac{sin2\alpha}{sin2\alpha}

∴ L.S = \frac{1}{sin2\alpha} + \frac{sin2\alpha}{sin2\alpha}

→ The denominators are equal, then add the numerators

∴ L.S = \frac{1+sin2\alpha}{sin2\alpha}

∵ R. S = \frac{(sin\alpha+cos\alpha)^{2} }{sin2\alpha}

∵ (sinα + cosα)² = sin²α + 2 sinα cosα + cos²α

∴ (sinα + cosα)² = sin²α + cos²α + 2 sinα cosα

→ By using the 1st rule above, equate sin²α + cos²α by 1

∴ (sinα + cosα)² = 1 + 2 sinα cosα

→ By using the 2nd rule above, equate 2 sinα cosα by sin2α

∴ (sinα + cosα)² = 1 + sin2α

→ Substitute it in the R.S above

∴ R. S = \frac{1+sin2\alpha}{sin2\alpha}

∵ L.S = R.S

∴ csc 2α + 1 = \frac{(sin\alpha+cos\alpha)^{2} }{sin2\alpha}

4 0
3 years ago
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