Using the slope and 1 point format: (y- y₁) = m(x - x₁)
where m = slope = 5, and point is (x₁, y₁). In this case point (x₁, y₁) = <span>(-2, -1)
</span><span>x₁ = -2, y₁ = -1
</span>
Using: (y- y₁)<span> = m(x </span><span>- x₁)
</span>
y - -1 = 5(x - -2)
y + 1 = 5(x + 2).
We can go further but since our options are in that format above, we stop there.
So answer is option B.
Jesus is always the answer
Answer:
the answer for this question 1200
Step-by-step explanation:
Answer:
1/ sqrt(1+ln^2(x)) * 1/(ln^2x +1) * 1/x
Step-by-step explanation:
f(x) = sin (tan^-1 (ln(x)))
u substitution
d/du (sin u) * du /dx
cos (u) * du/dx
Let u =(tan^-1 (ln(x))) du/dx =d/dx (tan^-1 (ln(x)))
v substitution
Let v = ln x dv/dx = 1/x
d/dv (tan ^-1 v) dv/dx
1/( v^2+1) * dv/dx
=1/(ln^2x +1) * 1/x
Substituting this back in for du/dx
cos (tan^-1 (ln(x)) * 1/(ln^2x +1) * 1/x
We know that cos (tan^-1 (a)) = 1/ sqrt(1+a^2)
cos (tan^-1 (ln(x)) * 1/(ln^2x +1) * 1/x
1/ sqrt(1+ln^2(x)) * 1/(ln^2x +1) * 1/x
