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mart [117]
3 years ago
11

How to solve the volume formula for cylinders using pi​

Mathematics
1 answer:
otez555 [7]3 years ago
6 0

The formula for the volume of a cylinder is V=Bh or V=πr2h . The radius of the cylinder is 8 cm and the height is 15 cm. Substitute 8 for r and 15 for h in the formula V=πr2h . i think the answer is this

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How many tiles would I need to by if each tile was 2ft by 2Ft and the square footage was 69X69
valina [46]
This is the formula you would use to calculate how many tiles you would need.

# of tiles you need = Total square footage ÷ square footage of each tile 


Total square footage: 69x69 = 4761 square feet
Square footage of each tile = 2x2 = 4 square feet

# of tiles you need =
4761 ÷ 4 = 1190.25 

You will need 1190.25 tiles.
3 0
3 years ago
What is the equation of the lime that passes through the point (1,3) and has a slope of 6
Ilya [14]

Answer:

y = 6x+ 3.

Step-by-step explanation:

y - y1 = m(x - x1)

sub.

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7 0
2 years ago
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Find the 73rd term of the arithmetic sequence -28, -41, -54, ...
PolarNik [594]

Answer:

the 73rd term of the arithmetic sequence is -964.

Step-by-step explanation:

The common difference in this arithmetic sequence is 13.  This is obtained by subtracting 28 from 41.  The first term is a(1) = -28.

The arithmetic sequence formula is a(n) = a(1) + d(n - 1), where d is the common difference and -28 the first term.

Then, in this case, a(n) = -28 - 13(n - 1), and

a(73) = -28 - 13(73 - 1) = -964

3 0
2 years ago
Hellppp plsss will hibe brainliest answer
WINSTONCH [101]

Answer:

try this answer: 72

5 0
2 years ago
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For questions 13-15, Let Z1=2(cos(pi/5)+i Sin(pi/5)) And Z2=8(cos(7pi/6)+i Sin(7pi/6)). Calculate The Following Keeping Your Ans
weqwewe [10]

Answer:

Step-by-step explanation:

Given the following complex values Z₁=2(cos(π/5)+i Sin(πi/5)) And Z₂=8(cos(7π/6)+i Sin(7π/6)). We are to calculate the following complex numbers;

a) Z₁Z₂ = 2(cos(π/5)+i Sin(πi/5)) * 8(cos(7π/6)+i Sin(7π/6))

Z₁Z₂ = 18 {(cos(π/5)+i Sin(π/5))*(cos(7π/6)+i Sin(7π/6)) }

Z₁Z₂ = 18{cos(π/5)cos(7π/6) + icos(π/5)sin(7π/6)+i Sin(π/5)cos(7π/6)+i²Sin(π/5)Sin(7π/6)) }

since i² = -1

Z₁Z₂ = 18{cos(π/5)cos(7π/6) + icos(π/5)sin(7π/6)+i Sin(π/5)cos(7π/6)-Sin(π/5)Sin(7π/6)) }

Z₁Z₂ = 18{cos(π/5)cos(7π/6) -Sin(π/5)Sin(7π/6) + i(cos(π/5)sin(7π/6)+ Sin(π/5)cos(7π/6)) }

From trigonometry identity, cos(A+B) = cosAcosB - sinAsinB and  sin(A+B) = sinAcosB + cosAsinB

The equation becomes

= 18{cos(π/5+7π/6) + isin(π/5+7π/6)) }

= 18{cos((6π+35π)/30) + isin(6π+35π)/30)) }

= 18{cos((41π)/30) + isin(41π)/30)) }

b) z2 value has already been given in polar form and it is equivalent to 8(cos(7pi/6)+i Sin(7pi/6))

c) for z1/z2 = 2(cos(pi/5)+i Sin(pi/5))/8(cos(7pi/6)+i Sin(7pi/6))

let A = pi/5 and B = 7pi/6

z1/z2 = 2(cos(A)+i Sin(A))/8(cos(B)+i Sin(B))

On rationalizing we will have;

= 2(cos(A)+i Sin(A))/8(cos(B)+i Sin(B)) * 8(cos(B)-i Sin(B))/8(cos(B)-i Sin(B))

= 16{cosAcosB-icosAsinB+isinAcosB-sinAsinB}/64{cos²B+sin²B}

= 16{cosAcosB-sinAsinB-i(cosAsinB-sinAcosB)}/64{cos²B+sin²B}

From trigonometry identity; cos²B+sin²B = 1

= 16{cos(A+ B)-i(sin(A+B)}/64

=  16{cos(pi/5+ 7pi/6)-i(sin(pi/5+7pi/6)}/64

= 16{ (cos 41π/30)-isin(41π/30)}/64

Z1/Z2 = (cos 41π/30)-isin(41π/30)/4

8 0
3 years ago
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