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Blizzard [7]
3 years ago
5

2y-8+4y what is equal to this please help thanks

Mathematics
1 answer:
Nitella [24]3 years ago
8 0

Answer:

6y - 8

Or

2(3y - 4)

Step-by-step explanation:

The given expresion is 2y-8+4y.

We want to find an expression that is equal to 2y-8+4y.

We need to group like terms to obtain:

2y + 4y - 8

We can now combine the like terms to get:

2y + 4y - 8

6y - 8

We can factor 2 to obtain:

2(3y - 4)

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Step-by-step explanation:

\frac{-x^{2}+9}{x+3} \\ \\ \frac{-(x^{2}-9)}{x+3} \\ \\ \frac{-(x-3)(x+3)}{x+3} \\ \\ -x(x-3)

4 0
2 years ago
If one million people have equal chances of being called by telephone from a broadcasting show, then the probability of one part
zhenek [66]
1 /million is the chance
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3 years ago
A line passes through the point (0, 2) and has a slope of -1/4 What is the equation of the line?
maw [93]

Answer:

The required equation is x + 4y = 8 !!

Step-by-step explanation:

<em><u>Given</u></em><em><u> </u></em><em><u>:</u></em><em><u>-</u></em><em><u> </u></em><em> </em><em>the</em><em> </em><em>line</em><em> </em><em>pass</em><em>es</em><em> </em><em>t</em><em>hrough</em><em> </em><em>the</em><em> </em><em>point</em><em> </em><em>(</em><em> </em><em>0</em><em> </em><em>,</em><em> </em><em>2</em><em> </em><em>)</em><em> </em><em>and</em><em> </em><em>the</em><em> </em><em>slope</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>line</em><em> </em><em>is</em><em> </em><em>(</em><em> </em><em>-</em><em>1</em><em>/</em><em>4</em><em> </em><em>)</em><em> </em>

<em>•</em><em> </em><em>Also</em><em>,</em><em> </em><em>to</em><em> </em><em>form</em><em> </em><em>an</em><em> </em><em>eq</em><em>uation</em><em> </em><em>when</em><em> </em><em>a</em><em> </em><em>po</em><em>int</em><em> </em><em>throu</em><em>gh</em><em> </em><em>which</em><em> </em><em>line</em><em> </em><em>passes</em><em> </em><em>and</em><em> </em><em>slope</em><em> </em><em>of</em><em> </em><em>line</em><em> </em><em>is</em><em> </em><em>given</em><em> </em><em>we</em><em> </em><em>use</em><em> </em><em>the</em><em> </em><em>formula</em><em> </em><em>;</em>

<em>(</em><em> </em><em>y</em><em> </em><em>-</em><em> </em><em>y1</em><em> </em><em>)</em><em> </em><em>=</em><em> </em><em>m</em><em> </em><em>(</em><em> </em><em>x</em><em> </em><em>-</em><em> </em><em>x1</em><em> </em><em>)</em>

<em>Where</em><em> </em><em>,</em><em> </em><em>y</em><em> </em><em>and</em><em> </em><em>x</em><em> </em><em>are</em><em> </em><em>vari</em><em>ables</em><em> </em>

<em>and</em><em> </em><em>(</em><em> </em><em>x1</em><em> </em><em>,</em><em> </em><em>y1 </em><em>)</em><em> </em><em>are</em><em> </em><em>the</em><em> </em><em>po</em><em>ints</em><em> </em><em>through</em><em> </em><em>which</em><em> </em><em>line </em><em>passes</em><em> </em>

<em>also</em><em>,</em><em> </em><em>m</em><em> </em><em>=</em><em> </em><em>slope</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>re</em><em>quired</em><em> </em><em>line</em><em> </em>

<em>Here</em><em> </em><em>,</em><em> </em><em>x1</em><em> </em><em>=</em><em> </em><em>0</em><em> </em><em>,</em><em> </em><em>y1</em><em> </em><em>=</em><em> </em><em>2</em><em> </em><em>and</em><em> </em><em>m</em><em> </em><em>=</em><em> </em><em>(</em><em> </em><em>-1</em><em>/</em><em>4</em><em> </em><em>)</em><em> </em>

<em>[</em><em> </em><em>Ref</em><em>er to</em><em> the</em><em> attached</em><em> file</em><em> for</em><em> </em><em>furth</em><em>er</em><em> </em><em>process</em><em> </em><em>]</em>

3 0
3 years ago
Write the slope intercept form of the equation of the line through the given points x intercept of -5 and y intercept of -1. Pls
soldier1979 [14.2K]

bearing in mind that an x-intercept is when the graph touches the x-axis and when that happens y = 0, and a y-intercept is  when the graph touches the y-axis and when that happens x = 0.

\bf \underset{x-intercept}{(\stackrel{x_1}{-5}~,~\stackrel{y_1}{0})}\qquad \underset{y-intercept}{(\stackrel{x_2}{0}~,~\stackrel{y_2}{-1})} ~\hfill \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-1}-\stackrel{y1}{0}}}{\underset{run} {\underset{x_2}{0}-\underset{x_1}{(-5)}}}\implies \cfrac{-1}{0+5}\implies -\cfrac{1}{5}

\bf \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-\stackrel{y_1}{0}=\stackrel{m}{-\cfrac{1}{5}}[x-\stackrel{x_1}{(-5)}] \\\\\\ y=-\cfrac{1}{5}(x+5)\implies y = -\cfrac{1}{5}x-1

4 0
3 years ago
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