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Taya2010 [7]
3 years ago
5

What is the intersection of the three sets: A = {1, 3, 5, 7, 9}, B = {1, 2, 3, 5, 7}, and C = {1, 2, 5, 8, 9}?

Mathematics
1 answer:
marysya [2.9K]3 years ago
7 0

ITS A. 1 AND 5 BECAUSE 1 AND 5 ARE IN ALL 3 SETS

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Please help me too in if sin w=-0.5 and w is a degree measure in the third quadrant, find w.
MissTica
Do you remember your unit circle? If sin ω = was -1/2, then it would be 7<span>π/6. If you're unfamiliar with the unit circle, we can derive it.

So, you know that sin is OPPOSITE/HYPOTENUSE, and it's in the third quadrant, where x and y would be negative. If sin </span>ω = -1/2, that means that ω = 1/sin*(-1/2), or sin^(-1)*(-1/2). Let's ignore the negative for now and plug sin^(-1)*(-1/2) into your calculator in radians. You get (1/6)π. But that's in Quadrant 1. We want it in Quadrant 3.

In one complete revolution, or 360°, there are 2π radians. That means, if you want to rotate it 180°, you need to add π to what you originally got.
π+(1/6)π=(7/6)π.

I highly recommend you memorize the unit circle if you haven't already, because you'll need it from Precalculus on.
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3 years ago
In ΔABC (m∠C = 90°), the points D and E are the points where the angle bisectors of ∠A and ∠B intersect respectively sides BC an
Airida [17]

This is a little long, but it gets you there.

  1. ΔEBH ≅ ΔEBC . . . . HA theorem
  2. EH ≅ EC . . . . . . . . . CPCTC
  3. ∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
  4. ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
  5. ∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
  6. ∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
  7. ΔDAC ≅ ΔDAG . . . HA theorem
  8. DC ≅ DG . . . . . . . . . CPCTC
  9. ∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
  10. ∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
  11. ∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
  12. ∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
  13. (∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
  14. ∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
  15. This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)
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