<em>The</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>5</em><em>5</em><em>3</em><em>8</em><em>.</em><em>9</em><em>6</em><em> </em><em>units</em><em>^</em><em>2</em>
<em>please</em><em> </em><em>see</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em> for</em><em> </em><em>full</em><em> solution</em><em> </em>
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<em>Good</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>
Answer:
300. In this case, it's very easy to do this percentage inflation. Since 20 is a multiple of 100, and we want to know the 100%, you just multiply the 20 by 5 to make it 100. Along with this, you multiply the 60 by 5 as well, which gives you 300.
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.
For Q3, you can subsitute x = 2 and y = 5 into the systems
-5(2) + 5 = -5
so it's a solution for the first system
-4(2)+2(5) = 2
it's also a solution for the second system