Find the slope of the given line in the following graph.

Mathematics

1 answer:

andre [41]2 years ago

3 0

Take one point from the cartesian plane.

I am taking:

(5,-5)
x= 5
y= -5

$\large \sf \frac{y}{x}$

$\large \sf \cancel\frac{ - 5}{5}$

$\sf \: m = - 1$

You might be interested in

If f(x)=2x+sinx and the function g is the inverse of f then g'(2)=

Alexxx [7]

$\bf f(x)=y=2x+sin(x)
\\\\\\
inverse\implies x=2y+sin(y)\leftarrow f^{-1}(x)\leftarrow g(x)
\\\\\\
\textit{now, the "y" in the inverse, is really just g(x)}
\\\\\\
\textit{so, we can write it as }x=2g(x)+sin[g(x)]\\\\
-----------------------------\\\\$

$\bf \textit{let's use implicit differentiation}\\\\
1=2\cfrac{dg(x)}{dx}+cos[g(x)]\cdot \cfrac{dg(x)}{dx}\impliedby \textit{common factor}
\\\\\\
1=\cfrac{dg(x)}{dx}[2+cos[g(x)]]\implies \cfrac{1}{[2+cos[g(x)]]}=\cfrac{dg(x)}{dx}=g'(x)\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}$

now, if we just knew what g(2) is, we'd be golden, however, we dunno

BUT, recall, g(x) is the inverse of f(x), meaning, all domain for f(x) is really the range of g(x) and, the range for f(x), is the domain for g(x)

for inverse expressions, the domain and range is the same as the original, just switched over

so, g(2) = some range value
that means if we use that value in f(x), f( some range value) = 2

so... in short, instead of getting the range from g(2), let's get the domain of f(x) IF the range is 2

thus 2 = 2x+sin(x)

$\bf 2=2x+sin(x)\implies 0=2x+sin(x)-2
\\\\\\
-----------------------------\\\\
g'(2)=\cfrac{1}{2+cos[g(2)]}\implies g'(2)=\cfrac{1}{2+cos[2x+sin(x)-2]}$

hmmm I was looking for some constant value... but hmm, not sure there is one, so I think that'd be it

5 0

2 years ago

Simplify <br>it was x there so u not confused

Mariulka [41]

Step-by-step explanation:

here's the solution: -

=》

$\frac{x {}^{6} }{ {x}^{4} } \times {x}^{2}$