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3 years ago

5

Write the equation in standard form. Then factor the left side of the equation 2x2=28-X

1 answer:

inn [45]3 years ago

6 0

The correct answer is B) (2x-7)(x+4)=0. If you were to FOIL (First, Outer, Inner, Last) the answer, you would get 2x^2+1x-28=0.
Step by step:
(2x-7)(x+4)=0
FOIL it.
2x^2+8x-7x-28=0
Combine the two middle terms.
2x^2+1x-28=0
Move -28 and 1x over to the right side of the equation.
2x^2=28-x

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I believe that the last one is correct because it is not a curve but a straight line, and it can't be the second one because the y value does not decrease as x increases

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The number represented by x is less than 7 and is evenly divisible by both 0.6 and 2.4

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2 years ago

Midpoint of the segment between the points (−5,13) and (6,4)

Leokris [45]

Answer: $(0.5,8.5)$

Step-by-step explanation:

We need to use the following formula to find the Midpoint "M":

$M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

Given the points (-5,13) and (6,4) can identify that:

$x_1=-5\\x_2=6\\\\y_1=13\\y_2=4$

The final step is to substitute values into the formula.

Therefore, the midpoint of the segment between the points (-5,13) and (6,4) is:

$M=(\frac{-5+6}{2},\frac{13+4}{2})\\\\M=(\frac{1}{2},\frac{17}{2})\\\\M=(0.5,8.5)$

6 0

2 years ago

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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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2 years ago

Read 2 more answers