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3 years ago

Hey I am struggling with this question and was hoping someone could help me before 7:00PM CST.

Mathematics

2 answers:

mariarad [96]3 years ago

4 0

$\bf \cfrac{5x}{x^2+2x}\div \cfrac{30x^2}{x+2}\implies \cfrac{5x}{x^2+2x}\cdot \cfrac{x+2}{30x^2}\implies \cfrac{\begin{matrix} 5x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}}{x~~\begin{matrix} (x+2) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}}\cdot \cfrac{\begin{matrix} x+2\\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix} }{\begin{matrix} 5x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}\cdot 6x} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \cfrac{1}{6x^2}~\hfill$

yKpoI14uk [10]3 years ago

3 0

Answer: x^3+30x+75

-----------------------------

15x

Step-by-step explanation:

5x/x^2+2x÷30x^2/x+2

1/15x^5+2x^3+5x^2

---------------------------

x^3

x^3+30x+75

-----------------------------

15x

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Answer:

The vertex of the quadratic function is:

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Step-by-step explanation:

Given the function

$f\left(x\right)=x^2+6x+16$

As the vertex of the form $y=ax^2+bx+c$ is defined as:

$x_v=-\frac{b}{2a}$

As the quadratic function of parabola params are

$a=1,\:b=6,\:c=16$

$x_v=-\frac{b}{2a}$

$x_v=-\frac{6}{2\cdot \:1}$

$x_v=-3$

Putting $x_v=-3$ to determine $y_v$

$y_v=\left(-3\right)^2+6\left(-3\right)+16$

$y_v=3^2-18+16$

$y_v=9-2$

$y_v=7$

Therefore, the vertex of the quadratic function is:

$(x_{v}, y_{v})=\left(-3,\:7\right)$

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3 years ago

Point A is on a number line halfway between −17−17 and 55. Point B is halfway between point A and 0. What integer does the point

julia-pushkina [17]

Find A first by using formula: (a+b)/2= midpoint
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3 years ago

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Answer:

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Step-by-step explanation:

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Answer:

170m

Step-by-step explanation:

The answer to the above question is letter d which is 170 m. To get the 170 m, kindly check the below solution:

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4 years ago

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-BARSIC- [3]

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