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pshichka [43]
4 years ago
9

Perform the indicated operation and simplify show work please:

Mathematics
2 answers:
NARA [144]4 years ago
3 0
\frac{x+4y}{x+y}+\frac{2x-y}{x+y}=\frac{x+4y+2x-y}{x+y}=\frac{3x+3y}{x+y}=\frac{3(x+y)}{x+y}=3\ ;(x \not= -y)
stepan [7]4 years ago
3 0
<u>x + 4y</u> + <u>2x - y </u>= <u>x + 2x + 4y - y</u> =<u> 3x + 3y</u> = 3
 x + y      x + y           x + y               x + y
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slamgirl [31]

Answer:

A bird drops a stick to the ground from a height of 80 feet. The equation 0=−16x2+64x+80 gives the number of seconds that have passed when it hits the ground. After about how many seconds did the stick hit the ground?

Step-by-step explanation:

h = -16t^2 + 80   Stick hits the ground when h = 0:

0 = -16t^2 + 80    

16t^2 = 80

t^2 = 80/16 = 5

t = POSITIVE root 5 (due to the real world nature of the problem)

t approximately 2.236 seconds.

========================================

Graph is a downward opening parabola with h-intercept of 80, t-intercept(s) of sqrt(5), - sqrt(5)

Symmetric about h-axis, though I would only graph on the right side of the h-axis since  

that is where t >= 0.

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3 years ago
The verticle of a triangle are (2, 18) (-2, -4), and (6,12.
Sergeu [11.5K]

so the triangle has the vertices of  (2, 18) (-2, -4), and (6,12), that gives us the endpoints for each line of

(2, 18) , (-2, -4)

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(6,12) , (2, 18)


\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{18})\qquad (\stackrel{x_2}{-2}~,~\stackrel{y_2}{-4}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-4-18}{-2-2}\implies \cfrac{-22}{-4}\implies \cfrac{11}{2} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-18=\cfrac{11}{2}(x-2) \\\\\\ y-18=\cfrac{11}{2}x-11 \implies \blacktriangleright y=\cfrac{11}{2}x+7 \blacktriangleleft


\bf (\stackrel{x_1}{-2}~,~\stackrel{y_1}{-4})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{12}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{12-(-4)}{6-(-2)}\implies \cfrac{12+4}{6+2}\implies \cfrac{16}{8}\implies 2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-4)=2[x-(-2)] \\\\\\ y+4=2(x+2)\implies y+4=2x+4\implies \blacktriangleright y=2x \blacktriangleleft


\bf (\stackrel{x_1}{6}~,~\stackrel{y_1}{12})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{18}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{18-12}{2-6}\implies \cfrac{6}{-3}\implies -2 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-12=-2(x-6) \\\\\\ y-12=-2x+12\implies \blacktriangleright y=-2x+24 \blacktriangleleft

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