if we take 2100 to be the 100%, what is 273 off of it in percentage?
<span>2 s; 22 ft 1 s; 22 ft 2 s; 6 ft 1 s; 54 ft i et itdont g</span>
Answer:
B
Step-by-step explanation:
The first step is to draw the line y = 3 - x to see what the line itself looks like. Is it going from left to right as in A and C is going up as you go from right to left as in B and D? The graph on the left (below) gives you the answer. It is going up as you progress from right to left.
The next step is to answer which is it: B or D.
y has to be above the line. it is B. See the graph below on the right.
Answer:
I'm taking an educated guess here and saying that its option two.
I haven't done these problems in almost two years.
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
