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Sphinxa [80]
2 years ago
8

Find the surface area of a cylinder with the height of 10 inches and a base diameter of 6 in ​

Mathematics
1 answer:
sweet-ann [11.9K]2 years ago
4 0

Answer:

339.24

Step-by-step explanation:

To find the surface area, you will need to the the area of the base of the cylinder. To do that, use the formula A=pi (r)^2

The radius is half the diameter do r=3

3.14(3)^2 should give you about 28.27.

This is the area of the top and bottom of the cylinder.

Now, do 28.27x10=282.7 to find the height surface area.

282.7+28.27+28.27=Surface area.

SA= about 339.24

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Answer: 170.37 acres

Step-by-step explanation:

In 2005, x= 2005

Substituting x=2005 in the equation.

A(2005) = (2078x-94458)/(13x-2164)

A(2005) = 4071932/23901

A(2005) = 170.37 acres

7 0
3 years ago
Read 2 more answers
(2 5/6)(6)+(-1 1/2)(1.5)-(1/16)
Jet001 [13]
Now I’m ngl my math might be wrong but I ended up with 1131/48
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4 0
3 years ago
Please help !!! :(<br> 1 and 2 form a linear pair, if the measure of 1 is 113 find the measure of 2
anzhelika [568]

Answer:

67

Step-by-step explanation:

A linear pair is two angles which form a straight angle when combined (or 180 degrees). Thus do 180 - 113 to get 67 degrees for Angle 2

6 0
3 years ago
What is the value of X to the nearest tenth?
Dennis_Churaev [7]

The answer to your question is 21.5

8 0
3 years ago
PLEASE HELP ASAP 25 PTS TO RIGHT/BEST ANSWER
Novay_Z [31]
Subtract 1111 from both sides

5{e}^{{4}^{x}}=22-115e​4​x​​​​=22−11


 

Simplify 22-1122−11 to 1111

5{e}^{{4}^{x}}=115e​4​x​​​​=11



 

Divide both sides by 55

{e}^{{4}^{x}}=\frac{11}{5}e​4​x​​​​=​5​​11​​



 

Use Definition of Natural Logarithm: {e}^{y}=xe​y​​=x if and only if \ln{x}=ylnx=y

{4}^{x}=\ln{\frac{11}{5}}4​x​​=ln​5​​11​​



 

: {b}^{a}=xb​a​​=x if and only if log_b(x)=alog​b​​(x)=a

x=\log_{4}{\ln{\frac{11}{5}}}x=log​4​​ln​5​​11​​



 

Use Change of Base Rule: \log_{b}{x}=\frac{\log_{a}{x}}{\log_{a}{b}}log​b​​x=​log​a​​b​​log​a​​x​​

x=\frac{\log{\ln{\frac{11}{5}}}}{\log{4}}x=​log4​​logln​5​​11​​​​



 

Use Power Rule: \log_{b}{{x}^{c}}=c\log_{b}{x}log​b​​x​c​​=clog​b​​x
\log{4}log4 -> \log{{2}^{2}}log2​2​​ -> 2\log{2}2log2

x=\frac{\log{\ln{\frac{11}{5}}}}{2\log{2}}x=​2log2​


Answer= −0.171
7 0
3 years ago
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