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Sliva [168]
3 years ago
9

Ignore circled one and work it out please​

Mathematics
1 answer:
mash [69]3 years ago
3 0

Answer:

a. \frac{(x-7)}{x^2+x-30}

Step-by-step explanation:

\huge \frac{x^2-3x-28}{(x^2+10x+24)(x-5)} \\\\\huge {=\frac{x^2-7x+4x-28}{(x^2+6x+4x+24)(x-5)}}\\\\\huge {=\frac{x(x-7)+4(x-7)}{\{x(x+6)+4(x+6)\}(x-5)}}\\\\\huge {=\frac{(x-7)(x+4)}{(x+6)(x+4)(x-5)}}\\\\\huge {=\frac{(x-7)}{(x+6)(x-5)}}\\\\\huge {=\frac{(x-7)}{x^2+(6-5)x+6(-5)}}\\\\\huge {=\frac{(x-7)}{x^2+x-30}}

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If HJ=2x+5, JK=3x-7, and KH=18 what is the value of x
weqwewe [10]

Answer:

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Step-by-step explanation:

<u>Given</u>

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<u>We see J is between H and K, so </u>

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3 0
3 years ago
Use addition or substitution to find the value of y for this set of equations. 12x - 5y = 30 y = 2x - 6
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Now that that is set up we can make our equation.
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We have to multiply the second equation by -6 in order to cancle out the X's. That comes out to (12x-6y=36)

Now the equation looks like (12x+5y=30) - (12x+6y=36)

After you subtract you have (-y=-6)

divide by (-1) and you're left with y=6

8 0
3 years ago
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Step-by-step explanation:

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3 years ago
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