A=7d+13
A=7(4)+13
A= 28+13
A=41
Answer:
See below.
Step-by-step explanation:
Since sqrt(a) and sqrt(b) are in simplest radical form, that means a and b have no perfect square factors. When sqrt(a) and sqrt(b) are multiplied giving c * sqrt(d), the fact that c came out of the root means that there was c^2 inside the product sqrt(ab). This means that a and b have at least one common factor.
ab = c^2d
Example:
Let a = 6 and let b = 10.
sqrt(6) and sqrt(10) are in simplest radical form.
Now we multiply the radicals.
sqrt(a) * sqrt(b) = sqrt(6) * sqrt(10) = sqrt(60) = sqrt(4 * 15) = 2sqrt(15)
We have c = 2 and d = 15.
ab = c^2d
6 * 10 = 2^2 * 15
60 = 60
Our relationship between a, b and c, d works.
<span>Answer:
Its too long to write here, so I will just state what I did.
I let P=(2ap,ap^2) and Q=(2aq,aq^2)
But x-coordinates of P and Q differ by (2a)
So P=(2ap,ap^2) BUT Q=(2ap - 2a, aq^2)
So Q=(2a(p-1), aq^2)
which means, 2aq = 2a(p-1)
therefore, q=p-1
then I subbed that value of q in aq^2
so Q=(2a(p-1), a(p-1)^2)
and P=(2ap,ap^2)
Using these two values, I found the midpoint which was:
M=( a(2p-1), [a(2p^2 - 2p + 1)]/2 )
then x = a(2p-1)
rearranging to make p the subject
p= (x+a)/2a</span>
Slope = Y2-Y1/X2-X1 = 7-(-3)/3-(-2)= 10/5 = 2
As is the case for any polynomial, the domain of this one is (-infinity, +infinity).
To find the range, we need to determine the minimum value that f(x) can have. The coefficients here are a=2, b=6 and c = 2,
The x-coordinate of the vertex is x = -b/(2a), which here is x = -6/4 = -3/2.
Evaluate the function at x = 3/2 to find the y-coordinate of the vertex, which is also the smallest value the function can take on. That happens to be y = -5/2, so the range is [-5/2, infinity).
