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3 years ago

-25 - 6k + 7=6(1+ 3k) Solve the equation for The Variable

Mathematics

2 answers:

dmitriy555 [2]3 years ago

6 0

Answer: Isolate the variable by dividing each side by factors that don't contain the variable.

k = −1

romanna [79]3 years ago

6 0

Answer:

-1

Step-by-step explanation:

-25+7 -6=6k+18k

-24=24k

k=-1

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Which of the following sets of lengths can represent the measures of the sides of a right triangle?

finlep [7]

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Explanation:
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4 years ago

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MA_775_DIABLO [31]

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3 years ago

Read 2 more answers

Divide a^24-a^18+a^12 by a^6<br> what is the quotation?

lana66690 [7]

I think you mean "what is the QUOTIENT?"

$\dfrac{a^{24} - a^{18} + a^{12}}{a^6} =$

$= \dfrac{a^{24}}{a^6} - \dfrac{a^{18}}{a^6} + \dfrac{a^{12}}{a^6}$

$= a^{24 - 6} - a^{18 - 6} + a^{12 - 6}$

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5 0

4 years ago

PLEASE PLEASE HELP ME OUT ITS URGENT <br><br><br> What are the solutions to the quadratic equation?

mixer [17]

Answer: Choice C) $x = 4\pm2\sqrt{10}$

This is the same as saying $x = 4+2\sqrt{10} \ \ \text{ or } \ \ x = 4-2\sqrt{10}$

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Explanation:

First get everything to one side

x^2 + 13 = 8x + 37

x^2 + 13 - 8x - 37 = 0

x^2 - 8x - 24 = 0

Here we have a quadratic in the form ax^2+bx+c = 0

Note how a = 1, b = -8, c = -24

Those values are plugged into the quadratic formula below

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x = \frac{-(-8)\pm\sqrt{(-8)^2-4(1)(-24)}}{2(1)}\\\\x = \frac{8\pm\sqrt{160}}{2}\\\\x = \frac{8\pm\sqrt{16*10}}{2}\\\\x = \frac{8\pm\sqrt{16}*\sqrt{10}}{2}\\\\x = \frac{8\pm4*\sqrt{10}}{2}\\\\x = \frac{2(4\pm2\sqrt{10})}{2}\\\\x = 4\pm2\sqrt{10}$