Answer:
#3: x is approximately 0.37, -5.37, #4: x is approximately 4.19, -1.19
Step-by-step explanation:
You can solve using the quadratic equation or by solving with the perfect square
= 
x is approximately 0.37, -5.37
Domain of f = R .............
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
G=3
If you multiply out x+7 and x-4 you get <span>x^2+3x-28
The 3 in the above equation is like the G</span>
Answer:
a) AB = 17 m
b) AC = 20.8 m
Step-by-step explanation:
Pythagorean theorem
a^2 + b^2 = c^2
a)
AB^2 = (26-11)^2 + 8^2
AB^2 = 15^2 + 8^2
AB^2 = 225 + 64
AB^2 = 289
AB = 17 m
b)
AC^2 = AB^2 + BC^2
AC^2 = 17^2 + 12^2
AC^2 =289 + 144
AC^2 = 433
AC = 20.8 m
