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4 years ago

12

Y=(x-4)(x+20) find vertex

1 answer:

andrew11 [14]4 years ago

7 0

Y = (x-4)(x+20) ===> 0 = (x-4)(x+20)

that means the roots of that quadratic are x = 4 and x = -20.

now, the vertex will be between these two guys, right in between.

from -20 to 4 there are 24 units, half-way of that is 12 units, so, if we move from -20, 12 units to the right or towards 4, we'll end up at -20 + 12 = -8.

so, the vertex's x-coordinate is -8 then, what's "y" anyway?

y = (x-4)(x+20) x = -8

y = [(-8) - 4] [(-8) + 20]

y = (-12) ( 12)

y = -144

therefore, the vertex is at (-8 , -144)

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Answer: 7.7 years

Step-by-step explanation:

Given

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3 0

3 years ago

<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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Read 2 more answers

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azamat

Answer:

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This is a one step solution

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