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Romashka-Z-Leto [24]
3 years ago
11

The diameter of a circle is 14 inches. Find the circumstance and area use 22/7

Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
3 0

Answer:

Circumference= 44 in

Area= 154 in²

Step-by-step explanation:

\boxed{circumference \: of \: circle = 2\pi  r}

Radius

= diameter ÷2

= 14 ÷2

= 7 in

Circumference of circle

= 2( \frac{22}{7} )(7)

= 44 in

\boxed{area \: of \: circle = \pi {r}^{2} }

Area of circle

=  \frac{22}{7} ( {7}^{2} )

= 154 in²

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What would the answer be
Hoochie [10]

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B.

Step-by-step explanation:

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A box at a miniature golf course contains contains 4 red golf balls, 8 green golf balls, and 7 yellow golf balls. What is the pr
Solnce55 [7]

Answer:

 =57.89%

Step-by-step explanation:

The total number of golf ball is 4+8+7 = 19

P (red or yellow) = number of red or yellow

                              ------------------------------------

                                total number of golf balls

                           = 4+7

                              -----

                              19

                         =11/19

Changing this to a percent means changing it to a decimal and multiplying by 100%

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Rounding to two decimal places

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Graph each number on a number line. Write an inequality that is true for the numbers. a.13/8 and 3/8
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3 years ago
Solve the equation.
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Answer:

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Step-by-step explanation:

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  0.1x +18.8 = -4 +2x

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  12 = x . . . . . . . . . . . . . divide by 1.9

__

2) Eliminate parentheses:

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  3.2x = 28.8 . . . . . . . . add 16 - 0.8x

  x = 9 . . . . . . . . . . . . . .divide by 3.2

_____

<em>Comments on the solutions</em>

The expression we add in each case eliminates the constant on one side of the equation and the variable term on the other side. That leaves an equation of the form ...

  variable term = constant

We choose to eliminate the smaller variable term (the one with the coefficient farthest to the left on the number line). Then the constant we eliminate is the on on the other side of the equation. This choice ensures that the remaining variable term has a positive coefficient, tending to reduce errors.

__

You can work these problems by methods that eliminate fractions. Here, the fractions are decimal values, so are not that difficult to deal with. In any event, it is good to be able to work with numbers in any form: fractions, decimals, integers. It can save some steps.

7 0
3 years ago
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