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kenny6666 [7]
3 years ago
12

Please help (the question is in the attached image)

Mathematics
1 answer:
Alex73 [517]3 years ago
5 0

Answer:

a.the length of midsegment of an equilateral triangle with side lengths of 12.5 cm.=12.5/2=6.25cm

b.

sinceUT is the perpendicular bisector of line segment AB, where T is on AB, find the length of AT given AT = 3x + 6 and TB = 42 - x.

than AT=BT

3x+6=42-x

3x+x=42-6

x=36/4=9

AT=3×9+6=27+6=33

c.

since angle EFG has angle bisector FH, where EF = GF, find the value of y if EH = 5y + 10 and HG = 28 - y.

EH=GH

5y+10=28-y

6y=28-10

y=18/6=3

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3 years ago
Evaluate the given integral by changing to polar coordinates. 8xy dA D , where D is the disk with center the origin and radius 9
BabaBlast [244]

Answer:

0

Step-by-step explanation:

∫∫8xydA

converting to polar coordinates, x = rcosθ and y = rsinθ and dA = rdrdθ.

So,

∫∫8xydA = ∫∫8(rcosθ)(rsinθ)rdrdθ = ∫∫8r²(cosθsinθ)rdrdθ = ∫∫8r³(cosθsinθ)drdθ

So we integrate r from 0 to 9 and θ from 0 to 2π.

∫∫8r³(cosθsinθ)drdθ = 8∫[∫r³dr](cosθsinθ)dθ

= 8∫[r⁴/4]₀⁹(cosθsinθ)dθ

= 8∫[9⁴/4 - 0⁴/4](cosθsinθ)dθ

= 8[6561/4]∫(cosθsinθ)dθ

= 13122∫(cosθsinθ)dθ

Since sin2θ = 2sinθcosθ, sinθcosθ = (sin2θ)/2

Substituting this we have

13122∫(cosθsinθ)dθ = 13122∫(1/2)(sin2θ)dθ

= 13122/2[-cos2θ]/2 from 0 to 2π

13122/2[-cos2θ]/2 = 13122/4[-cos2(2π) - cos2(0)]

= -13122/4[cos4π - cos(0)]

= -13122/4[1 - 1]

= -13122/4 × 0

= 0

5 0
3 years ago
Pls help me with this :)<br><br><br> .
san4es73 [151]

Answer:

35 is the answer thank u 35 is a real answer

3 0
3 years ago
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