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sattari [20]
3 years ago
6

Find the Area of the figure below, composed of a rectangle and a semicircle. Round to the nearest tenths place.

Mathematics
2 answers:
luda_lava [24]3 years ago
7 0

Answer:

286.97

Step-by-step explanation:

find the area of the rectangle first

14*15=210

Next

Find the area of a circle

πr^2=a

the radius of the circle is half the height of the rectangle

π*(7*7)=a

a=153.94

Now

Divide the area by two

153.94/2=76.97

Now add the two areas

76.97+210

Luden [163]3 years ago
5 0

Answer:

Area of the figure= area of the rectangle+ area of the semicircle

=(15×14)+(1/2)π(7)²=210+77=287 square unit

<u>287 square unit</u> is the right answer.

Remember:--- area of the rectangle

=(length×breadth)

and area of the semicircle=π(radius)²/2

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\dfrac{5x^4-7x^3-12x^2+6x+21}{(x-3)(x^2-2)^2}=\dfrac{a_1}{x-3}+\dfrac{a_2x+a_3}{x^2-2}+\dfrac{a_4x+a_5}{(x^2-2)^2}
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When x=3, you're left with

147=49a_1\implies a_1=\dfrac{147}{49}=3

When x=\sqrt2 or x=-\sqrt2, you're left with

\begin{cases}17-8\sqrt2=(\sqrt2a_4+a_5)(\sqrt2-3)&\text{for }x=\sqrt2\\17+8\sqrt2=(-\sqrt2a_4+a_5)(-\sqrt2-3)\end{cases}\implies\begin{cases}-5+\sqrt2=\sqrt2a_4+a_5\\-5-\sqrt2=-\sqrt2a_4+a_5\end{cases}

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