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2 years ago

The radius of a circle is 10 m. Find its area to the nearest whole number.

Mathematics

1 answer:

Bess [88]2 years ago

7 0

<h3>given:</h3>

radius= 10 m

the area of the given circle.

<h3>solution:</h3>

$a = \pi {r}^{2}$

$a = \pi \times {10}^{2}$

$a = 314.15927 \: {m}^{2}$

$a = 314 \: {m}^{2}$

therefore, the area of the given circle is 314 square meters. 

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Given each of the following areas, what would be the radius of each circle? a. A = 530.66 m2 b. A = 907.46 sq. ft. c. A = 379.94

ra1l [238]

Area for a circle is $A= \pi r^2$ . We will fill in each area and then solve for the radius. $530.66m^2= \pi r^2$ and $\frac{530.66m^2}{3.14}=r^2$ . ∴ $169=r^2$ and r = 13. For b, $907.46ft^2= \pi r^2$ and $\frac{907.46ft^2}{3.14}=r^2$ . $289=r^2$ , ∴ r = 17. For c, $379.94mi^2= \pi r^2$ , and $\frac{379.94mi^2}{3.14}=r^2$ . ∴, $121=r^2$ and r = 11. For d, $1962.5cm^2= \pi r^2$ , and $\frac{1962.5cm^2}{3.14}=r^2$ . ∴, $625=r^2$ so r = 25. And there you have it!

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3 years ago

12x+7<−11 AND5x−8≥4012

Arte-miy333 [17]

Answer:

No solution

Step-by-step explanation:

Given: $12x+7$ and $5x-8\geq 40$

Solve both inequality separately.

$12x+7$ Subtraction property of inequality

$12x$

$x$

and

$5x-8\geq 40$ Addition property of inequality

$5x\geq 40+8$

$5x\geq 48$

$x\geq \dfrac{48}{5}$

Hence, the solution of the compound inequality intersection of both solutions.

Please find attachment for number line solution.

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Please help 30 points Asap

Andrej [43]

40 units2˛

Looking at the figure, the rectangle has the vertexes (2,1), (3,-3), (-5,-5) and (-6,-1). The parallelogram has the vertexes (2,7), (3,3), (3,-3), and (2,1).

The area of a parallelogram is base times height. We have 2 vertical lines at x=2 and x=3, so the height is 1. And the length of the line from (3,3) to (3,-3) is 6, so the base is 6. Therefore the area of the parallelogram is 1*6 = 6.

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