Answer:
The integrals was calculated.
Step-by-step explanation:
We calculate integrals, and we get:
1) ∫ x^4 ln(x) dx=\frac{x^5 · ln(x)}{5} - \frac{x^5}{25}
2) ∫ arcsin(y) dy= y arcsin(y)+\sqrt{1-y²}
3) ∫ e^{-θ} cos(3θ) dθ = \frac{e^{-θ} ( 3sin(3θ)-cos(3θ) )}{10}
4) \int\limits^1_0 {x^3 · \sqrt{4+x^2} } \, dx = \frac{x²(x²+4)^{3/2}}{5} - \frac{8(x²+4)^{3/2}}{15} = \frac{64}{15} - \frac{5^{3/2}}{3}
5) \int\limits^{π/8}_0 {cos^4 (2x) } \, dx =\frac{sin(8x} + 8sin(4x)+24x}{6}=
=\frac{3π+8}{64}
6) ∫ sin^3 (x) dx = \frac{cos^3 (x)}{3} - cos x
7) ∫ sec^4 (x) tan^3 (x) dx = \frac{tan^6(x)}{6} + \frac{tan^4(x)}{4}
8) ∫ tan^5 (x) sec(x) dx = \frac{sec^5 (x)}{5} -\frac{2sec^3 (x)}{3}+ sec x
Let's say is "x".
so, "x" is the whole or 10/10 whilst 0.002 is 1/10 then, let's check
Answer:
y=27 and x=9
Step-by-step explanation:
the ratio is 4 to 3
First, find the slope of the line y-Ax=B
y=Ax+B, so the slope is A, which is a constant.
Because the new line is parallel to the given line, so the slope is also A.
Next, write the new line in point-slope form:
(y-3)=A(x-(-7)) =>y-3=A(x+7)
You can write it into slope-intercept form if needed: y-3=Ax+7A =>y=Ax+7A+3
or standard form: y-Ax-7A-3=0
331 I think, I could be wrong tho
