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NemiM [27]
3 years ago
10

Which of the following represents a relation that is not a function?

Mathematics
1 answer:
Doss [256]3 years ago
7 0
A function is for every input, there is 1 output
aka
x never repeats with a different y

A. x doesn't repeat, it's a function

B. -8 repeats with 34 and 40, this is not a funciton (this is the answer0

C. x doesn't repeat, it's a function

D. x doesn't repeat, it's a function



answer is B
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Answer:

y = 3/4x + 6

Step-by-step explanation:

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------- =  ------

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y = mx + b

take coordinates for y and x out of (-4, 3) or (0,6)

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7 0
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The answer is 7 i think
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I did the first question already I need help with the second pls :>
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Answer:

The box is considered a cube so,

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3 years ago
Vanderbilt Corporation, which utilizes a calendar year as its fiscal year, is the payee on a $12,000, 7%, 8-month note receivabl
Vedmedyk [2.9K]
Given:
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Notes Receivable from Stanford Company
<span>$12,000, 7%, 8-month 
                               
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3 0
3 years ago
How do you find a vector that is orthogonal to 5i + 12j ?
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\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{a}{b}\\\\&#10;slope=\cfrac{a}{{{ b}}}\qquad negative\implies  -\cfrac{a}{{{ b}}}\qquad reciprocal\implies - \cfrac{{{ b}}}{a}\\\\&#10;-------------------------------\\\\

\bf \boxed{5i+12j}\implies &#10;\begin{array}{rllll}&#10;\ \textless \ 5&,&12\ \textgreater \ \\&#10;x&&y&#10;\end{array}\quad slope=\cfrac{y}{x}\implies \cfrac{12}{5}&#10;\\\\\\&#10;slope=\cfrac{12}{{{ 5}}}\qquad negative\implies  -\cfrac{12}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{12}&#10;\\\\\\&#10;\ \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.

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\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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