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3 years ago

7

Solve by completing the square: 5x^2 + 20x +32 = 0 (please show work step by step)

1 answer:

marusya05 [52]3 years ago

7 0

5[x²+(2)²—(2)²]+32

5[(x+2)²—4]+32

5(x+2)²—20+32

5(x+2)²+12

Ans

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Given f(x)=x^4-7x^3+6x^2+8x+9

posledela

The workings to the answers are in the attachments below. Open them up in a new window to see them in full.

Lowest Point On The Graph:

(4.480878208, -61.32441280)

Highest Point On The Graph:

(1.155422854, 17.23818241)

2nd Lowest Point On The Graph:

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8 0

3 years ago

Which of the following represents the zeros of f(x) = 2x3 − 5x2 − 28x + 15?

likoan [24]

$\mathrm{Use\:the\:rational\:root\:theorem}$

$a_0=15,\:\quad a_n=2$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:3,\:5,\:15,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\:2$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:3,\:5,\:15}{1,\:2}$

$-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3$

$-\frac{3}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+3$

$\mathrm{Compute\:}\frac{2x^3-5x^2-28x+15}{x+3}\mathrm{\:to\:get\:the\:rest\:of\:the\:eqution:\quad }2x^2-11x+5$

$=\left(x+3\right)\left(2x^2-11x+5\right)$

$Factor: 2x^2-11x+5$

$2x^2-11x+5=\left(2x^2-x\right)+\left(-10x+5\right)$

$=x\left(2x-1\right)-5\left(2x-1\right)$

$2x^3-5x^2-28x+15=\left(x+3\right)\left(x-5\right)\left(2x-1\right)$

$\left(x+3\right)\left(x-5\right)\left(2x-1\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:}$

thus zeros of f(x) is

$x=-3,\:x=5,\:x=\frac{1}{2}$

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The sum of two complementary angles is 90°. For one pair of complementary angles, the measure of the first angle is 15 less than

steposvetlana [31]

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1st angle = 35 degrees

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Answer:

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Step-by-step explanation:

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