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kati45 [8]
3 years ago
6

One solution to 3x+9+4x+x

Mathematics
2 answers:
stira [4]3 years ago
7 0

Answer:

8x+9

Step-by-step explanation:

Katyanochek1 [597]3 years ago
5 0

Answer:

8x+9

Step-by-step explanation:

Add the like terms together.

3x+9+4x+x = 8x+9

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<em><u>9</u></em><em><u>0</u></em><em><u>. </u></em><em><u>(</u></em><em><u> </u></em><em><u>was </u></em><em><u>the </u></em><em><u>full </u></em><em><u>capacity</u></em><em><u> of</u></em><em><u> </u></em><em><u>jug</u></em><em><u>)</u></em>

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<em><u> </u></em><em><u> </u></em>

<em><u>be</u></em><em><u>. </u></em><em><u>x</u></em>

<em><u>so,</u></em><em><u> </u></em>

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Answer:

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Step-by-step explanation:

wkanda might never be forever which means that still could be quite ttue?

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Answer:

Step-by-step explanation:

\displaystyle\  \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\frac{0}{0} \\\\we\ can \ use\ Hospital's\ Rule\\\\\\f(x)=\sqrt{2x}-\sqrt{3x-a}  \qquad  f'(x)=\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} \\\\g(x)=\sqrt{x} -\sqrt{a}  \qquad g'(x)=\dfrac{1}{2\sqrt{x}} \\\\\\\displaystyle\  \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\lim_{n \to a} \dfrac{\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}}  }{\dfrac{1}{2\sqrt{x}} }\\\\

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