The standard eqn of a parabola in vertex form is y-k = a(x-h)^2, where (h,k) is the vertex. There are a good number of steps involved. I don't think it wise not to "show work." I cannot answer this question without going through all those steps.
However, there's an easier way to find the vertex. Identify the coefficients a, b and c:
a= -4, b= -3 and c = 1
Then the x-coord. of the vertex is x = -b / (2a). Subst. -3 for b and -4 for a and simplify. x = ??
Then find the y-coord. of the vertex by subbing your result, above, into the original equation.
Write the vertex as (h,k).
Once you have this vertex, you can find the equation in vertex form as follows:
Start with the general form y-k = a(x-h)^2, where (h,k) is the vertex.
You've already found the vertex (h,k). Subst. h and k into the general form, above. Then only the coefficient "a" remains undefined.
The answer to the question is-4x+524z+10
Answer is •D because of the seburation
Answer:
0.03%
Step-by-step explanation:
12(2a-3)=-4a+5
Remember to follow PEMDAS. Note the equal sign, what you do to one side, you do to the other.
First, distribute 12 to all terms within the parenthesis
12(2a) = 24a
12(-3) = -36
24a - 36 = -4a + 5
Add 4a to both sides, and add 36 to both sides
24a (+4a) - 36 (+36) = -4a (+4a) + 5 (+36)
24a + 4a = 5 + 36
Simplify
28a = 41
Isolate the a. Divide 28 from both sides
28a/28 = 41/28
a = 41/28
Simplify a
a = 1.46 (simplified)
hope this helps
