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2 years ago

If the sum of a number and three is doubled?, the result is seven less than the number. Find the number

Mathematics

1 answer:

pentagon [3]2 years ago

6 0

(x+3)*2=x-7
2x+6=x-7
x=-13

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<img src="https://tex.z-dn.net/?f=%5Csqrt%7B%28-81%29x%5E%7B2%7D%20%7D" id="TexFormula1" title="\sqrt{(-81)x^{2} }" alt="\sqrt{(

SSSSS [86.1K]

Answer:

We conclude that:

$\sqrt{\left(-81\right)x^2}=9ix$

Step-by-step explanation:

Given the radical expression

$\sqrt{\left(-81\right)x^2}$

simplifying the expression

$\sqrt{\left(-81\right)x^2}$

Remove parentheses: (-a) = -a

$\sqrt{\left(-81\right)x^2}=\sqrt{-81x^2}$

Apply radical rule: $\sqrt{-a}=\sqrt{-1}\sqrt{a},\:\quad \mathrm{\:assuming\:}a\ge 0$

$=\sqrt{-1}\sqrt{81x^2}$

Apply imaginary number rule: $\sqrt{-1}=i$

$=i\sqrt{81x^2}$

Apply radical rule: $\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0$

$=\sqrt{81}i\sqrt{x^2}$

$=9i\sqrt{x^2}$

Apply radical rule: $\sqrt[n]{a^n}=a,\:\quad \mathrm{\:assuming\:}a\ge 0$

$=9ix$

Therefore, we conclude that:

$\sqrt{\left(-81\right)x^2}=9ix$

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2 years ago

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