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3 years ago

How do you rewrite 6^4×6^4 as a single power?

Mathematics

2 answers:

tensa zangetsu [6.8K]3 years ago

6 0

Answer:

6^8

Step-by-step explanation:

6^4 * 6^4

We know that a^b* a^c = a^(b+c)

6^4×6^4 = 6^(4+4) = 6^8

kondor19780726 [428]3 years ago

6 0

Answer:

$6^8$

Step-by-step explanation:

<u>Multiplication of powers with same base is addition in the powers.</u>

<u /> $6^4*6^4$

6^(4+4)

$6^8$

Answer: $6^8$

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What numbers? Be specific.

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Read 2 more answers

Which of the following are solutions to the equation below?

sladkih [1.3K]

Answer:

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=3,\:x=-\frac{1}{2}$

Step-by-step explanation:

considering the equation

$2x^2\:-\:4x\:-\:3\:=\:x$

solving

$2x^2\:-\:4x\:-\:3\:=\:x$

$2x^2-4x-3-x=x-x$

$2x^2-5x-3=0$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=2,\:b=-5,\:c=-3:\quad x_{1,\:2}=\frac{-\left(-5\right)\pm \sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

solving

$x=\frac{-\left(-5\right)+\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$x=\frac{5+\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$x=\frac{5+\sqrt{49}}{2\cdot \:2}$

$x=\frac{5+7}{4}$

$x=3$

also solving

$x=\frac{-\left(-5\right)-\sqrt{\left(-5\right)^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$x=\frac{5-\sqrt{\left(-5\right)^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$x=\frac{5-\sqrt{49}}{4}$

$x=-\frac{2}{4}$

$x=-\frac{1}{2}$

Therefore,

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=3,\:x=-\frac{1}{2}$

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3 years ago

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Answer:

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Step-by-step explanation:

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