Find the circumference of a cd that has an diamater of 5 round your andwer to the nereast tenth

Mathematics

1 answer:

aev [14]2 years ago

7 0

Answer:

15.7

Step-by-step explanation:

C = 2πr

C = 2(3.14)(2.5)

C = 15.7

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Solve the equation

irina1246 [14]

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X=-9

6M-5=1
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5A+7=4A
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3 years ago

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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.

ΔEBH ≅ ΔEBC . . . . HA theorem
EH ≅ EC . . . . . . . . . CPCTC
∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
ΔDAC ≅ ΔDAG . . . HA theorem
DC ≅ DG . . . . . . . . . CPCTC
∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
(∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
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