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Wittaler [7]
3 years ago
7

Which answer is right?

Mathematics
1 answer:
AleksandrR [38]3 years ago
5 0
the answer to this problem is (1,3)
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Gora finished 23% of his project last week, and another 32% of the entire project this week.What fraction of the project dose he
sergiy2304 [10]

Answer: Mark me brainliest..

Step-by-step explanation: and imma put the answer inside the comments if you do so. youve got nothing to lose! :D

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What is 14/28 in simplest form?
zhenek [66]
Hi!

The GCM of 14 and 28 is 7. 7 goes into 14 2 times, and goes into 28 4 times. So...

2/4

Which can be simplified further to 1/2

The answer is 1/2

Hope this helps! :)
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Is this answer proven to be correct dear sir?
lukranit [14]
Dear sir, we know not what you mean.
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in the 2004 United stated presidential election, 11.6 million we on were old enough to vote. of those 60% voted. out of 98.5 mil
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Just multiply the numbers by decimals and add the original number to what you got from multiplying and then subtract the two

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4 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).)
5 0
3 years ago
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