EXPLANATION
Let's see the facts:
The scale is---> 14 milimeters ----> 5 meters
Width = 42 milimeters
Applying the unitary method, the actual width of the pool is:
The width of the pool is 15m
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
Answer: 90º
Step-by-step explanation: complimentary angles are either of the two angles whose sum is 90º
The best and most correct answers among the choices provided by the question are:
<span>▢A. -2x+3y=-15
▢B. y=2/3x-5</span>
y = mx + b. where m is the slope<span> of the line and b is the y-</span>intercept<span> of the line, or the y-coordinate of the point at which the line crosses the y-axis. To write an equation in </span>slope-intercept form<span>, given a graph of that equation, pick two points on the line and use them to find the </span>slope<span>.</span>
I hope my answer has come to your help. God bless and have a nice day ahead!
Answer:
put one number in front of the other
