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MAVERICK [17]
3 years ago
6

how do you use equations to solve problems about area of rectangles parallelograms trapezoids and triangles

Mathematics
1 answer:
Bogdan [553]3 years ago
8 0
Area of parellelogram and rectangle=bh

trapezoid A=(b1+b2)h÷2

triangle A=bh÷2
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Solve the triangle.
saw5 [17]

Answer:

Step-by-step explanation:

because <u>c</u> is shorter than<u> a</u> we know C is a smaller angle than A

so the first choice is out.

use the law of cosines to find b

b^{2} = a^{2} +c^{2} -2*a*c*cos(B)

b^{2} =1681+400-1326.787871

b^{2}= 754.21219

b= \sqrt{754.21219}

b= 27.46292

so the last choice looks good

5 0
3 years ago
What is 4.338 as a fraction when the 338 is repeating<br>i give brianliest
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If <em>x</em> = 4.338338338…, then

1000<em>x</em> = 4338.338338338…

and subtracting <em>x</em> from this eliminate the trailing decimal.

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999<em>x</em> = 4334

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6 0
2 years ago
A pencil has a mass of 25 grams. An apple a mass that is 75 grams more than the pencil has What is the mass of the apple, in gra
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The apple weighs 100 grams 
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What is the product? 2x+8 x2 + 3x-4 x(x-4)(x-1) 20x+4) 2 (x+4)(x-4) 2x(x-1) 2x(x+4)​
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=−640x10+1280x9+19904x8−40728x7−144488x6+323904x5−162304x4+1024x3+2048x2

step by step

(2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x(x+4)

=((2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x)(x+4)

=((2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x)(x)+((2x+8x2+3x−4x(x−4)(x−1)(20)x+4)(2)(x+4)(x−4)(2)x(x−1)(2)x)(4)

=−640x10+3840x9+4544x8−58904x7+91128x6−40608x5+128x4+512x3−2560x9+15360x8+18176x7−235616x6+364512x5−162432x4+512x3+2048x2

=−640x10+1280x9+19904x8−40728x7−144488x6+323904x5−162304x4+1024x3+2048x2

7 0
3 years ago
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