Answer:
By hypotenuse - side test (HL) the two triangles are congruent.
Step-by-step explanation:
In ∆ABC and ∆DCB
i) angleABC = angleDCB.....(each 90°)
ii) BC = BC .....( common side)
iii) AC = DB.....(given)
therefore by hypo-side test ∆ABC congruent ∆DCB
The circumference of the circle is = 2 x pi x 6 = 37.7cm
To find the arc = 360 divided by 169 = 2.13
The arc = 37.7 divided by 2.13 = 17.69 cm
Answer:
The picture isnt working
Step-by-step explanation:
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
Think these are the answers to the first few hope it helps
