question seems incomplete but with the limited information one can deduce that the professor was trying to investigate how students CGPA is affected by the length of period spent on video games.

Step-by-step exp

independent variables are constants. they do not change or depend on any other variable.

dependent variables, they do depend on other variables, we call them resulting variables..

5 0

3 years ago

What are the x-intercepts of the equation? <img src="https://tex.z-dn.net/?f=r%28x%29%20%3D%20x-1%2Fx%2B4" id="TexFormula1"

cestrela7 [59]

Answer:

The only x-intercept is x = 1

Step-by-step explanation:

The equation is:

$r(x)=\frac{x-1}{x+4}$

We can substitute y for r(x), to write in the notation:

$r(x)=\frac{x-1}{x+4}\\y=\frac{x-1}{x+4}$

To get y-intercept, we put x = 0

and

To get x-intercept, we put y =0

We want to find x-intercepts here, so we substitute 0 into y and solve for x. Shown below:

$y=\frac{x-1}{x+4}\\0=\frac{x-1}{x+4}\\0(x+4)=x-1\\0=x-1\\x=1$

The only x-intercept is x = 1

3 0

3 years ago

Let g(x) = -5/?. Find g(g(-1))

IRISSAK [1]

Answer:

-1

Step-by-step explanation:

g(-1) = -5/-1 = 5

g(5) = -5/5

g(g(-1)) = -1

7 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

$\bf \textit{unit vector}\qquad \cfrac{\ \textless \ a,b\ \textgreater \ }{||\ \textless \ a,b\ \textgreater \ ||}\implies \cfrac{\ \textless \ a,b\ \textgreater \ }{\sqrt{a^2+b^2}}\implies \cfrac{a}{\sqrt{a^2+b^2}},\cfrac{b}{\sqrt{a^2+b^2}}
\\\\\\
\cfrac{12,-5}{\sqrt{12^2+5^2}}\implies \cfrac{12,-5}{13}\implies \boxed{\cfrac{12}{13}\ ,\ \cfrac{-5}{13}}
\\\\\\
\cfrac{-12,5}{\sqrt{12^2+5^2}}\implies \cfrac{-12,5}{13}\implies \boxed{\cfrac{-12}{13}\ ,\ \cfrac{5}{13}}$

4 0

3 years ago