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
1)
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
#1
"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:
A=(1/2)(b)(h)
sub in the new values to get A=1/2(8)(6) = 24
#2
vertical asymptotes are caused by the denominator becoming zero
look at the denominator, it factors to
(x+3)(x-1)
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
. So the equation of the vertical asymptotes would be x=-3 and x=1.
To find the equation of any vertical asymptotes for the function f(x) = (2x + 3)/(x^2 + 2x - 3), we need to analyze the denominator of the expression.
The denominator is x^2 + 2x - 3. To find the values of x that make the denominator equal to zero, we need to solve the equation x^2 + 2x - 3 = 0.
We can factor the equation as (x + 3)(x - 1) = 0.
Setting each factor equal to zero, we get:
x + 3 = 0 --> x = -3
x - 1 = 0 --> x = 1
So, the vertical asymptotes for the graph of f(x) are x = -3 and x = 1.
To determine the vertical asymptotes of a rational function, you need to find the values of x that make the denominator equal to zero.
In the given equation f(x) = (2x+3)/(x^2+2x-3), set the denominator equal to zero and solve for x:
x^2 + 2x - 3 = 0
Now, factorize the quadratic equation:
(x+3)(x-1) = 0
Setting each factor equal to zero gives us:
x+3 = 0 or x-1 = 0
Solving for x in each equation yields:
x = -3 or x = 1
So, the vertical asymptotes of the graph of f(x) = (2x+3)/(x^2+2x-3) are x = -3 and x = 1.