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Sedbober [7]
3 years ago
7

Work out the area of this circle radius 10cm take pie to be 3.142 and write down all the digest given by your calculator

Mathematics
2 answers:
Crazy boy [7]3 years ago
8 0

Answer:

314.2 cm²

Step-by-step explanation:

A = π·r²

= 3.142·10²

= 3.142·100

= 314.2 cm²

Alika [10]3 years ago
7 0

Answer:

314.2

Step-by-step explanation:

Area of a circle is: πr²

if radius is 10 then A = π*10² = 3.142 *100 =

314.2

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Answer: 12 ÷ 1 3/5

The problem is asking us to find how many days food would last. Since the food shelter uses 1 3/5 each day, the 12 bags of food would be split into 1 3/5 they use each day.

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3 years ago
4. Describe the proportion method for solving a percentage problem:
kondor19780726 [428]

Step-by-step explanation:

Consider the provided information.

For the proportion method first set up the equation like this:

\frac{Part}{Whole}=\frac{Percentage}{100}

Perform the cross multiplication and then solve for the missing part.

For example:

Find 80 percentage of 10.

Step 1: Set up the equation.

\frac{Part}{10}=\frac{80}{100}

\frac{Part}{10}=\frac{4}{5}

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Part=\frac{4}{5}\times 10

Step 3: Solve for the missing part.

Part=4 \times 2

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3 0
3 years ago
Read 2 more answers
Use the graph that shows the soulution to f(x)=g(x).
Kipish [7]

Step-by-step explanation:

The solutions are the points of intersection of f(x) and g(x), which are x = 0 or x = 2.

3 0
3 years ago
A gym's membership cost $280 in 2019. Now, it is $160. Which expression equals the percent of change?
kherson [118]

Answer:

1/2%

Step-by-step explanation:

4 0
3 years ago
Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis. Verify y
mariarad [96]

Answer:

the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis is;

\frac{\pi }{2}  [e^2 - 1 ]  or 10.036

Step-by-step explanation:

Given the data in the question;

y = y = e^{(x - 1 ), y = 0, x = 1, x = 2.

Now, using the integration capabilities of a graphing utility

y = y = e_2}^{(x - 1 )_, y = 0

Volume = \pi \int\limits^2_1 ( e^{x-1)^2} - (0)^2 dx

Volume = \pi \int\limits^2_1 ( e^{x-1)^2  dx

Volume = \pi \int\limits^2_1 e^{2x-2}dx

Volume = \frac{\pi }{e^2} \int\limits^2_1 e^{2x}dx

Volume = \frac{\pi }{e^2}  [\frac{e^{2x}}{2}]^2_1

Volume = \frac{\pi }{2e^2}  [e^4 - e^2 ]  

Volume = \frac{\pi }{2}  [e^2 - 1 ]  or 10.036

Therefore, the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis is;

\frac{\pi }{2}  [e^2 - 1 ]  or 10.036

   

3 0
3 years ago
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