The standard equation for a parabola that opens upward parallel to the y-axis is x2=4ay, where a = distance from the vertex to focus. Vertex is at origin(0,0)
4a=2
a=2/4
a=1/2 or 0.5
the coordinate of the focus is (0, 1/2)
the equation of the directrix is y=-1/2
Answer:
x=
3
2
=0.667
x=3
Step-by-step explanation:
I have taken that test (although I don't see you're statements)
I believe the statements to choose from are:
A.) The slope of the line is −10.
B.) The slope of the line is 3.
C.) One point on the line is (3, 6).
D.) One point on the line is (3,−6)
<u>The answers are:</u>
A.) The slope of the line is -10
D.) One point on the line is (3,-6)
<u>Explanation: </u>
The given equation of line is (1). The point slope form of a line is (2) Where m is the slope of line and (x₁,y₁) are points. On comparing (1) and (2) we get The slope of given line is -10 and the line passing through the points (3,-6).
We are given :

Step 1: factor the part inside square root
The function given inside square root is of quadratic form.
So let us try to factorise it using AC method.
Here A*C = 4*25 = 100
so we have to find factors of 100 that add up to give -20.
the two factors are -10 and -10.
Rewriting the function :

=
=
=
Step 2:
Now we take square root of the factorised form

= 
Answer : (2x-5)
Answer: D) cube root of 16
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Explanation:
The rule we use is
![x^{m/n} = \sqrt[n]{x^m}](https://tex.z-dn.net/?f=x%5E%7Bm%2Fn%7D%20%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D)
In this case, x = 4, m = 2 and n = 3.
So,
![x^{m/n} = \sqrt[n]{x^m}\\\\\\4^{2/3} = \sqrt[3]{4^2}\\\\\\4^{2/3} = \sqrt[3]{16}\\\\\\](https://tex.z-dn.net/?f=x%5E%7Bm%2Fn%7D%20%3D%20%5Csqrt%5Bn%5D%7Bx%5Em%7D%5C%5C%5C%5C%5C%5C4%5E%7B2%2F3%7D%20%3D%20%5Csqrt%5B3%5D%7B4%5E2%7D%5C%5C%5C%5C%5C%5C4%5E%7B2%2F3%7D%20%3D%20%5Csqrt%5B3%5D%7B16%7D%5C%5C%5C%5C%5C%5C)
Showing that the original expression turns into the cube root of 16.