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zaharov [31]
3 years ago
11

Will give BRAINLIEST Find the area of each circle . Round to the nearest tenth.

Mathematics
1 answer:
Oksi-84 [34.3K]3 years ago
3 0

Answer:

13:  312yd squared

14: 907.5 rounded up but not rounded is 907.46

Step-by-step explanation:

this is because the formula is pi*r^2  so you'd

Step by step:

1. 20yd/2= 10 to get the radius for 13

2. substitute r in the formula pi*r^2 for 10 making it pi*10^2

3. multiply 10 by itself, 10*10=100

4. multiply pi by 100, 3.14*100= 314

314yd^2 is the answer for 13

14:

1. 34ft/2=17 to get the radius for 14

2. substitute r on the formula pi*r^2 for 17 making it pi*17^2

3. multiply 17 by itself, 17*17=289

4. multiply pi by 289, 3.14*289=907.46

5. round 907.46 to the nearest tenth, 907.5

907.5ft^2 is your answer

hope this helps :)

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The coordinates of rhombus ABCD are A(–4, –2), B(–2, 6), C(6, 8), and D(4, 0). What is the area of the rhombus? Round to the nea
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Check the picture below.

so the rhombus has the diagonals of AC and BD, now keeping in mind that the diagonals bisect each, namely they cut each other in two equal halves, let's find the length of each.

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
A(\stackrel{x_1}{-4}~,~\stackrel{y_1}{-2})\qquad 
C(\stackrel{x_2}{6}~,~\stackrel{y_2}{8})\qquad \qquad 
%  distance value
d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}
\\\\\\
AC=\sqrt{[6-(-4)]^2+[8-(-2)]^2}\implies AC=\sqrt{(6+4)^2+(8+2)^2}
\\\\\\
AC=\sqrt{10^2+10^2}\implies AC=\sqrt{10^2(2)}\implies \boxed{AC=10\sqrt{2}}\\\\
-------------------------------

\bf ~~~~~~~~~~~~\textit{distance between 2 points}
\\\\
B(\stackrel{x_1}{-2}~,~\stackrel{y_1}{6})\qquad 
D(\stackrel{x_2}{4}~,~\stackrel{y_2}{0})\qquad \qquad BD=\sqrt{[4-(-2)]^2+[0-6]^2}
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BD=\sqrt{(4+2)^2+(-6)^2}\implies BD=\sqrt{6^2+6^2}
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BD=\sqrt{6^2(2)}\implies \boxed{BD=6\sqrt{2}}

that simply means that each triangle has a side that is half of 10√2 and another side that's half of 6√2.

namely, each triangle has a "base" of 3√2, and a "height" of 5√2, keeping in mind that all triangles are congruent, then their area is,

\bf \stackrel{\textit{area of the four congruent triangles}}{4\left[ \cfrac{1}{2}(3\sqrt{2})(5\sqrt{2}) \right]\implies 4\left[ \cfrac{1}{2}(15\cdot (\sqrt{2})^2) \right]}\implies 4\left[ \cfrac{1}{2}(15\cdot 2) \right]
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4[15]\implies 60

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Find the average value of f over region
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