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Darina [25.2K]
3 years ago
6

HELPPP!! Triangle ABC is similar to triangle MOP. A = 45 and P = 102. What is the measure of B?

Mathematics
1 answer:
erma4kov [3.2K]3 years ago
6 0

Answer:

       \large\boxed{\large\boxed{33\º}}

Explanation:

The order of the letters naming the triangles is relevant. When it is said that <em>triangle ABC is similar to triangle MOP</em>, means that:

  • ∡A ≅ ∡M
  • ∡B ≅ ∡O
  • ∡C ≅ ∡P

Therefore:

  • ∡A = 45º (given)
  • ∡P = 102º (given)
  • ∡C = 102º (by substitution)

Now, you have two angles of triangle ABC and can find the third one:

  • ∡A + ∡B + ∡C = 180º
  • ∡B = 180º - ∡A - ∡B
  • ∡B = 180º - 45º - 102º = 33º ← answer

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3 years ago
_ 5x^20-7x^10+15 in quadratic form
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Answer:

x = -(sqrt(349) - 7)^(1/10)/10^(1/10) or x = (sqrt(349) - 7)^(1/10)/10^(1/10) or x = -((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -(1/10 (-7 - sqrt(349)))^(1/10) or x = (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(1/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(1/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(2/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(2/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(3/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(3/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(4/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(4/5) (1/10 (-7 - sqrt(349)))^(1/10)

Step-by-step explanation:

Solve for x:

-5 x^20 - 7 x^10 + 15 = 0

Substitute y = x^10:

-5 y^2 - 7 y + 15 = 0

Divide both sides by -5:

y^2 + (7 y)/5 - 3 = 0

Add 3 to both sides:

y^2 + (7 y)/5 = 3

Add 49/100 to both sides:

y^2 + (7 y)/5 + 49/100 = 349/100

Write the left hand side as a square:

(y + 7/10)^2 = 349/100

Take the square root of both sides:

y + 7/10 = sqrt(349)/10 or y + 7/10 = -sqrt(349)/10

Subtract 7/10 from both sides:

y = sqrt(349)/10 - 7/10 or y + 7/10 = -sqrt(349)/10

Substitute back for y = x^10:

x^10 = sqrt(349)/10 - 7/10 or y + 7/10 = -sqrt(349)/10

Taking 10^th roots gives (sqrt(349)/10 - 7/10)^(1/10) times the 10^th roots of unity:

x = -(1/10 (sqrt(349) - 7))^(1/10) or x = (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(1/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(1/5) (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(2/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(2/5) (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(3/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(3/5) (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(4/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(4/5) (1/10 (sqrt(349) - 7))^(1/10) or y + 7/10 = -sqrt(349)/10

Subtract 7/10 from both sides:

x = -(sqrt(349) - 7)^(1/10)/10^(1/10) or x = (sqrt(349) - 7)^(1/10)/10^(1/10) or x = -((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or y = -7/10 - sqrt(349)/10

Substitute back for y = x^10:

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Taking 10^th roots gives (-7/10 - sqrt(349)/10)^(1/10) times the 10^th roots of unity:

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Answer:

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Tax equals the original amount times the tax rate

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