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horrorfan [7]
3 years ago
11

Write an equation in​ slope-intercept form of the line that passes through the given point and is parallel to the graph of the g

iven equation. ​(0,0); y=3/7x+ 7
Mathematics
1 answer:
AleksAgata [21]3 years ago
8 0

Answer:

y=3/7x

Step-by-step explanation:

Parallel means same slope.

y-y1=m(x-x1)

y-0=3/7(x-0)

y=3/7(x)+0

y=3/7x

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12:8:100= 3:2:25 is the simplest form
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Can some help please
Katyanochek1 [597]

Answer:

Step-by-step explanation:

2 * 10^7   You are to use a single digit. That's the 2 on the left. Then count what it takes to get the decimal between the 2 and the 3. It's 7

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3 years ago
How much more area does a large pizza with a 12in. diameter have than a small pizza with a 8 in. diameter? Round your answer to
USPshnik [31]

Answer:

A is the correct answer.

Step-by-step explanation:

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6 0
3 years ago
How do you find a vector that is orthogonal to 5i + 12j ?
Rashid [163]
\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\
slope=\cfrac{a}{{{ b}}}\qquad negative\implies  -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\
-------------------------------\\\\

\bf \boxed{5i+12j}\implies 
\begin{array}{rllll}
\ \textless \ 5&,&12\ \textgreater \ \\
x&&y
\end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5}
\\\\\\
slope=\cfrac{12}{{{ 5}}}\qquad negative\implies  -\cfrac{12}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{12}
\\\\\\
\ \textless \ 12, -5\ \textgreater \ \ or\ \ \textless \ -12,5\ \textgreater \ \implies \boxed{12i-5j\ or\ -12i+5j}

if we were to place <5, 12> in standard position, so it'd be originating from 0,0, then the rise is 12 and the run is 5.

so any other vector that has a negative reciprocal slope to it, will then be perpendicular or "orthogonal" to it.

so... for example a parallel to <-12, 5> is say hmmm < -144, 60>, if you simplify that fraction, you'd end up with <-12, 5>, since all we did was multiply both coordinates by 12.

or using a unit vector for those above, then

\bf \textit{unit vector}\qquad \cfrac{\ \textless \ a,b\ \textgreater \ }{||\ \textless \ a,b\ \textgreater \ ||}\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{\sqrt{a^2+b^2}}\implies \cfrac{a}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}&#10;\\\\\\&#10;\cfrac{12,-5}{\sqrt{12^2+5^2}}\implies \cfrac{12,-5}{13}\implies \boxed{\cfrac{12}{13}\ ,\ \cfrac{-5}{13}}&#10;\\\\\\&#10;\cfrac{-12,5}{\sqrt{12^2+5^2}}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}
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Find the value of x. Then classify the triangle
vredina [299]
So basically what you have to do is
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