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Irina18 [472]
3 years ago
7

How you would find the total area of the following composite shapes.

Mathematics
2 answers:
sergiy2304 [10]3 years ago
7 0

Answer:

By breaking it up in shapes

Step-by-step explanation:

I annotated your photo and it's attached below :) that's an example on how to break shapes into easier parts. So yeah

Download pdf
Elden [556K]3 years ago
6 0

Answer:

First one is 7.1415926,

Second one is 84 inch^{2}

Step-by-step explanation:

First for the square plus the two half circles: The two half circles are semi-circles, so two semi-circles put together equals a full circle.

Equation for the circle:

\pi (2/2)^{2} \\=\pi (1)^{2}

=\pi

≅3.1415926 inch^{2} (This is too long for me to write down, actually no body can write it down , since it's an infinity unrepeating decimal)

Then the square's area is:

2^{2} \\=4

Lastly 4+3.1415926

=7.1415926

Second one I am going to draw in the file first, then I'll write how I solved it here!

OK, so:

(5)(10)+(7)(4)+(15-(5+7))(4)/2)\\=50+28+(12/2)\\=78+6\\=84 inch^{2}

HOPED I helped!

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Answer: This was a bit hard to understand, x times the 5th root of x

Step-by-step explanation:

When you multiply square roots of the same root and inside value, they essentially get rid of the square roots. So the first two terms boil down to just x. Then multiply x by the 5th root of x to get:

x\sqrt[5]{x}

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At this market,6 batteries cost 10.38 how much do 8 batteries cost
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3 years ago
PLEASE HELP ME! Find the cube roots of 125(cos 288° + i sin 288°).
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◆ COMPLEX NUMBERS ◆

125 ( cos 288 + i sin 288 )  can be written as -

125.e^i( 288) 
125.e^i( 288 +360 ) 
125.e^i( 288+ 720)

[ As , multiples of 360 can be added to an angle without changing any trigonometric functions or sign ]

To find the cube root , take the cube root of above 3 expressions ,

We get -
5 e^( i 96 ) 
5 e^( i 216 )  
5 e^( i 336 )

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antiseptic1488 [7]

Answer:

\sf -11+7\sqrt{2}

Step-by-step explanation:

Given expression:

\sf \dfrac{3-\sqrt{32}}{1+\sqrt{2} }

Rewrite 32 as 16 · 2:

\sf \implies \dfrac{3-\sqrt{16 \cdot 2}}{1+\sqrt{2} }

Apply radical rule \sf \sqrt{a \cdot b}=\sqrt{a}\sqrt{b}

\sf \implies \dfrac{3-\sqrt{16}\sqrt{2}}{1+\sqrt{2} }

As \sf \sqrt{16}=4:

\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} }

Multiply by the conjugate:

\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} } \times \dfrac{1-\sqrt{2} }{1-\sqrt{2} }

\sf \implies \dfrac{(3-4\sqrt{2})(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}

\sf \implies \dfrac{3-3\sqrt{2}-4\sqrt{2}+4\sqrt{2}\sqrt{2}}{1-\sqrt{2}+\sqrt{2}-\sqrt{2}\sqrt{2}}

As \sf \sqrt{2}\sqrt{2}=\sqrt{4}=2:

\sf \implies \dfrac{3-3\sqrt{2}-4\sqrt{2}+4 \cdot 2}{1-\sqrt{2}+\sqrt{2}-2}

\sf \implies \dfrac{3-7\sqrt{2}+8}{1-2}

\sf \implies \dfrac{11-7\sqrt{2}}{-1}

\sf \implies -11+7\sqrt{2}

7 0
2 years ago
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