Answer:
No they are not equivalent
Answer:
Isosceles
Step-by-step explanation:
Consider triangle ABC, BD is the median and the altitude drawn to the side AC. This segment (BD) divide the triangle ABC into two triangles: ABD and CBD.
In these triangles:
- AD=DC (because BD is the median);
- ∠ADB=∠CDB=90° (because BD is the altitude);
- BD is common side.
Thus, by SAS postulate, triangles ABD and CBD are conruent. Congruent triangles have congruent corresponding sides. Hence, AB=CB.
If in triangle ABC, AB=BC, then this triangle is isosceles.
It's A.
Rae made the error when she added 7.
the equation should be:
<span>-14 - 7 = 7x + 7 - 7 </span>
Answer:
No solution
Step-by-step explanation:
21−7x=−7x−21−4
21−7x=−7x−25
21=−25
No solution
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
