Post a New Question


posted by on .

1)The area A of a triangle varies jointly as the lengths of its base b and height h. If A is 75 when b=15 and h=10,find A when b=8 and h=6

2)Determine the equation of any vertical asymtotes of the graph of f(x)=2x+3/x^2+2x-3

Area of a triangle is 1/2 b*h=A
use the formula and if you have any trouble post your work and I'll see If I can help.

24? thanks for all your help

1) What they are saying is that the area is proptional to the product b*h. If bh decreases by a factor 48/150, the area increases by the same factor. The smaller triangle area is
(48/150)*75 = 24
You could just as easily used the formula for the area of any triangle, (1/2) b h, but they didn't want you to do it that way for some reason. Blame your state's math curriculum committee.
(2) Asymptotes are straight lines that you approach when x becomes very large, whether positive or negative. The way you have written the equation, the 3/x^2 term becomes negligile at large |x|, and the equation becomes y = 4x - 3. That asymptote is not vertical. Is there supposed to be more terms than x^2 in the denominator following 3/ ?. If so, please indicate the denominator with parentheses

"The area A of a triangle varies jointly as the lengths of its base b and height h" means Area = k(b)(h)

so you are given Area=75, b=15, h=10 gets you k=1/2

so now we know the area of a triangle formula:
sub in the new values to get A=1/2(8)(6) = 24


vertical asymptotes are caused by the denominator becoming zero
look at the denominator, it factors to
so it becomes zero when x=-3 or x=1
so the equations of the verical asymtotes are
x=-3, x=1

I don't know. The top is 2x+3 and the bottom is x^2+2x-3

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question