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Genrish500 [490]
2 years ago
9

Evaluate 4+8 divided by2x (6-3)

Mathematics
1 answer:
Masja [62]2 years ago
8 0

Answer:

if the "x" is a multiplication symbol, the answer is 2.

Step-by-step explanation:

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The lengths of the sides of a triangle are in the ratio
Anton [14]
HOW TO SOLVE:

1) use x as a variable
2) the starting equation is:
          4x + 3x + 5x = 18

3) add all the like terms on the left side of the equation:
          12x = 18

4) Divide by 12 on both sides:
          x=1.5

5) Find the lengths of the sides:

        - 4x = 4(1.5) = 6

        - 3x = 3(1.5) =  4.5

       - 5x = 5(1.5) = 7.5

SO, the lengths of the sides of this triangle are: 6, 4.5, and 7.5
5 0
3 years ago
Determine the intercepts of the function f(x)=2x^2+18x+28
Sonja [21]
The intercepts are -2 and -7 on the x-axis
7 0
3 years ago
Solve x2 + 10x = 24 by completing the square. Which is the solution set of the equation? {–12, 2} {–2, 12}
guajiro [1.7K]

Answer: I could be wrong so don't get your hopes up but I believe the solution set would be {-2,12}

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
(1, 5) (3 , 1) <br> Please Help
hammer [34]

Answer:

(4, 6)?

Step-by-step explanation:

Im not sure you didn't give much info

6 0
2 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}

5 0
2 years ago
Read 2 more answers
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