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3 years ago

Which of the values in the set {2, 3, 4, 5} is a solution to the equation 2x + 4 = 10?

Mathematics

2 answers:

Mrac [35]3 years ago

7 0

The answer is 3 because 2(3) + 4 = 10; 6 + 4 = 10

svetoff [14.1K]3 years ago

7 0

Answer:

Step-by-step explanation:

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If the perimeter of a square is 5 feet now many Inches long is each side of the square

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Well since we know that the perimeter of a square is four times the length of one of its sides. We just have to divide 5 by 4 to get the length of one side:

5feet /4 sides = 1.25 feet

And to finish off, we have to convert feet to inches:

1 foot = 12 inches

1.25 feet x inches

x inches = 12 inches x 1.25 feet ÷ 1 foot

x inches = 15 inches

Therefore, each side is 15 inches long.

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

likoan [24]

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

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

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

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

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

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

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

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

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

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

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

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

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

thus zeros of f(x) is

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

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