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3 years ago

ABCD is a rectangle. AD = 15, AC = 25, and DC = 20. Find BD.

Mathematics

1 answer:

agasfer [191]3 years ago

6 0

Answer:

Step-by-step explanation:

it is the same length as DC just going diagonal.

(i am not good at explaining in case you could not tell)

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3 years ago

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2 years ago

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What is 40-35/35×100

leva [86]

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3 years ago

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1. 8y^2-4y+1 divided by 2y-1

____ [38]

Q1. The answer is $4y+ \frac{1}{2y-1}$

$\frac{8y^{2}-4y+1 }{2y-1} = \frac{4y*2y-4y+1}{2y-1} = \frac{4y(2y-1)+1}{2y-1} = \frac{4y(2y-1)}{2y-1}+ \frac{1}{2y-1} =4y+ \frac{1}{2y-1}$

Q2. The answer is $2a+3+ \frac{6}{a-1}$

$\frac{2 a^{2}+a+3 }{a-1} = \frac{a*2a-2a+3a+3}{a-1} = \frac{2a(a-1)+3a+3}{a-1}= \frac{2a(a-1)}{a-1}+ \frac{3a+3}{a-1} = \\ \\ =2a+ \frac{3a+3}{a-1}=2a+ \frac{3a-3+3+3}{a-1}=2a+ \frac{3(a-1)+6}{a-1} =2a+ \frac{3(a-1)}{a-1} + \frac{6}{a-1}= \\ \\ =2a+3+ \frac{6}{a-1}$


Q3. The answer is $2 x^{2} +5x+2$ 

 $\frac{6 x^{3} +11 x^{2} -4x-4}{3x-2} = \frac{3x*2 x^{2}-2 x^{2} *2+15 x^{2} -4x-4 }{3x-2} = \frac{2 x^{2} (3x-2)+15 x^{2} -4x-4}{3x-2}= \\ \\ = \frac{2 x^{2} (3x-2)}{3x-2} + \frac{15 x^{2} -4x-4}{3x-2} =2 x^{2} +\frac{15 x^{2} -4x-4}{3x-2}=2 x^{2} + \frac{15 x^{2} -10x+6x-4}{3x-2}= \\ \\ =2 x^{2} + \frac{5x*3x-5x*2+6x-4}{3x-2} =2 x^{2} + \frac{5x(3x-2)+3x*2-2*2}{3x-2} = \\ \\ =2 x^{2} + \frac{5x(3x-2)}{3x-2} + \frac{3x*2-2*2}{3x-2} =2 x^{2} +5x+ \frac{2(3x-2)}{3x-2} =2 x^{2} +5x+2$


Q4. The answer is 2x + 7

 $\frac{6 x^{2} +11x-35}{3x-5} = \frac{6 x^{2} -10x+21x-35 }{3x-5} = \frac{3 x *2x-5*2x+7*3x-7*5 }{3x-5} = \\ \\ = \frac{2x(3x-5)+7(3x-5)}{3x-5}= = \frac{(3x-5)(2x+7)}{3x-5} =2x+7$


Q5. The answer is $x+1- \frac{3}{x-1}$ 

 $\frac{ x^{2} -4}{x-1} = \frac{ x^{2} -x+x-1-3 }{x-1} = \frac{x*x-x+x-1-3}{x-1} = \frac{x(x-1)+(x-1)-3}{x-1} = \\ \\ \frac{(x+1)(x-1)-3}{x-1} = \frac{(x+1)(x-1)}{x-1} -\frac{3}{x-1} =x+1- \frac{3}{x-1}$

Q6. The answer is $y^{2} -2y+3$

$\frac{ y^{3}-4 y^{2}+7y-6 }{y-2} = \frac{y* y^{2} -2y^{2}-2 y^{2} +7y-6 }{y-2} = \frac{y^{2}(y-2)-2 y^{2} +7y-6}{y-2}= \\ \\ = \frac{y^{2}(y-2)}{y-2}+ \frac{-2 y^{2} +7y-6}{y-2} = y^{2} + \frac{-2 y^{2} +4y + 3y-6}{y-2} = \\ \\ =y^{2} + \frac{-2y*y-2y(-2)+3y-3*2}{y-2} = y^{2} + \frac{(-2y)(y-2)+3(y-2)}{y-2} = \\ \\ = y^{2} + \frac{(-2y+3)(y-2)}{y-2} = y^{2} +(-2y+3) =y^{2} -2y+3$


Q7. The answer is $x^{2} +xy+ y^{2}}{x-y}$ 

 $\frac{ x^{3} - \frac{x}{y} y^{3} }{x-y} = \frac{(x-y)( x^{2} +xy+ y^{2}) }{x-y} = \frac{ x^{2} +xy+ y^{2}}{x-y}$


Q8. The answer is $(a^{2} +2ab+2b^{2})$ 

 $\frac{a^{4} +4b^{4} }{a^{2}-2ab+2 b^{2} } = \frac{ (a^{2})^{2} +(2b)^{2}}{a^{2}-2ab+2 b^{2}} = \frac{(a^{2} -2ab+2b^{2})(a^{2} +2ab+2b^{2}) }{(a^{2} -2ab+2b^{2})} =(a^{2} +2ab+2b^{2})$ 

Q9. The answer is a^{n-8} - a^{-14}

 $\frac{ (a^{2}) ^{n} - a^{n-6} }{ a^{n+8} }= \frac{(a^{2}) ^{n} }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}} \\ \\ (x^{y}) ^{z}= x^{y*z} \\ \\ \frac{ x^{y} }{ x^{z} } = x^{y-z} \\ \\ 
\frac{(a^{2}) ^{n} }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}} =\frac{a^{2n} }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}} = a^{2n-(n+8)} - a^{n-6-(n+8)} = \\ \\ =a^{2n-n-8} - a^{n-6-n-8} = a^{n-8} - a^{-14}$

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3 years ago

What is the length of the midsegment of the trapezoid

makkiz [27]

Answer:

The length of the mid-segment of the trapezoid = 7

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Step-by-step explanation:

The mid-segment of a trapezoid is the segment that connecting the midpoints of the two non-parallel sides.

As shown in the figure the two non-parallel sides are AB and CD

∴ The mid-segment of the trapezoid = $\frac{BC +AD}{2}$

From the figure: BC = 8 and AD = 6

∴ The mid-segment of the trapezoid = $\frac{8+6}{2} = 7$

7 0

3 years ago