Perimeter of square is P = 4s.....s = length of 1 side
so ur function rule is : y = 4x
17) f(x) = 16/(13-x).
In order to find domain, we need to set denominator expression equal to 0 and solve for x.
And that would be excluded value of domain.
13-x =0
Adding x on both sides, we get
13-x +x = x.
13=x.
Therefore, domain is All real numbers except 13.
18).f(x) = (x-4)(x+9)/(x^2-1).
In order to find the vertical asymptote, set denominator equal to 0 and solve for x.
x^2 -1 = 0
x^2 -1^2 = 0.
Factoring out
(x-1)(x+1) =0.
x-1=0 and x+1 =0.
x=1 and x=-1.
Therefore, Vertical asymptote would be
x=1 and x=-1
19) f(x) = (7x^2-3x-9)/(2x^2-4x+5)
We have degrees of numberator and denominator are same.
Therefore, Horizontal asymptote is the fraction of leading coefficents.
That is 7/2.
20) f(x)=(x^2+3x-2)/(x-2).
The degree of numerator is 2 and degree of denominator is 1.
2>1.
Degree of numerator > degree of denominator .
Therefore, there would no any Horizontal asymptote.
(Fog) = f(g(x))
g(-2)=-2+4=2
f(2)=2^2 +3 = 4+3 =7
Answer:
sin(x) and cos(y) are equal and have ratio of 1 : 1
First find the missing hypotenuse using Pythagoras theorem:
a² + b² = c²
12² + 5² = c²
c = √144+25
c = 13
using sine rule:
using cosine rule:
