9 hundred thousands equal how many tens
2 answers:
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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)
We have
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)
Answer:
x = 28
Step-by-step explanation:
If the quadrilaterals are similar, there is a proportionality among their sides:
The top side in the large figure (70) is to the top side in the small figure (10) in the same ratio as the left side (x) in the large figure is to the left side in the small figure (4). This in math terms is written as:
We can then solve for the unknown "x" by multiplying both sides by 4:
1) / sin x / = y ≥ 0 ;
We solve the equation
a = 1 ; b = 1 ; c = - 2 ;
y1 = ( - 1 + 3 ) / 2 = 1 ≥ 0 ( correct ) ;
y2 = ( - 1 - 3 ) / 2 = - 2 ≥ 0 ( false ) ;
Then, / sin x / = 1 ;
sinx = + 1 or sin x = - 1 ;
x ∈ { 90 } U { 270 } ;
x ∈ {90 , 270}.
We solve the equation
a = 1 ; b = 1 ; c = - 2 ;
y1 = ( - 1 + 3 ) / 2 = 1 ≥ 0 ( correct ) ;
y2 = ( - 1 - 3 ) / 2 = - 2 ≥ 0 ( false ) ;
Then, / sin x / = 1 ;
sinx = + 1 or sin x = - 1 ;
x ∈ { 90 } U { 270 } ;
x ∈ {90 , 270}.