I think it’s going to be 60
Answer:
see explanation
Step-by-step explanation:
Using the tangent ratio in the right triangle
tan A =
=
=
, then
∠ A =
(
) ≈ 41° ( to the nearest degree )
The sum of the angles in the triangle = 180° , then
∠ B + 41° + 90° = 180°
∠ B + 131° = 180° ( subtract 131° from both sides )
∠ B = 49°
Using Pythagoras' identity in the right triangle
AB² = BC² + AC² = 7² + 8² = 49 + 64 = 113 ( take square root of both sides )
AB =
≈ 10.6 ( to the nearest tenth )
Step-by-step explanation:
⇒12)It is an arithmetic sequence.
d=2-1=3-2=4-3=1
a(n) = a +(n-1)d
a(n) = 1+(n-1)1
The next three terms:
a(6) = 1+(6-1)1=6
a(7) = 1+(7-1)1=7
a(8) = 1+(8-1)1=8
⇒13)It is an arithmetic sequence.
d=0-3=-3-0=-6+3=-3
a(n) = a +(n-1)d
a(n) = 3+(n-1)-3
The next three terms:
a(5) = 3+(5-1)-3=-9
a(6) = 3+(6-1)-3=-12
a(7) = 3+(7-1)-3=-15
⇒14)It is <u>not </u>an arithmetic sequence.
⇒15) a(50) = 10 +(50-1)5
=<u>255</u>
<u>I hope this helps</u>
<u />
1. Domain.
We have

in the denominator, so:
![x^2-2x-3\neq0\\\\(x^2-2x+1)-4\neq0\\\\(x-1)^2-4\neq0\\\\(x-1)^2-2^2\neq0\qquad\qquad[\text{use }a^2-b^2=(a-b)(a+b)]\\\\(x-1-2)(x-1+2)\neq0\\\\ (x-3)(x+1)\neq0\\\\\boxed{x\neq3\qquad\wedge\qquad x\neq-1}](https://tex.z-dn.net/?f=x%5E2-2x-3%5Cneq0%5C%5C%5C%5C%28x%5E2-2x%2B1%29-4%5Cneq0%5C%5C%5C%5C%28x-1%29%5E2-4%5Cneq0%5C%5C%5C%5C%28x-1%29%5E2-2%5E2%5Cneq0%5Cqquad%5Cqquad%5B%5Ctext%7Buse%20%7Da%5E2-b%5E2%3D%28a-b%29%28a%2Bb%29%5D%5C%5C%5C%5C%28x-1-2%29%28x-1%2B2%29%5Cneq0%5C%5C%5C%5C%0A%28x-3%29%28x%2B1%29%5Cneq0%5C%5C%5C%5C%5Cboxed%7Bx%5Cneq3%5Cqquad%5Cwedge%5Cqquad%20x%5Cneq-1%7D)
So there is a hole or an asymptote at x = 3 and x = -1 and we know, that answer B) is wrong.
2. Asymptotes:

We have only one asymptote at x = -1 (and hole at x = 3), thus the correct answer is A)