Y1 is the simplest parabola. Its vertex is at (0,0) and it passes thru (2,4). This is enough info to conclude that y1 = x^2.
y4, the lower red graph, is a bit more of a challenge. We can easily identify its vertex, which is (-4,0), and several points on the grah, such as (2,-3).
Let's try this: assume that the general equation for a parabola is
y-k = a(x-h)^2, where (h,k) is the vertex. Subst. the known values,
-3-(-4) = a(2-0)^2. Then 1 = a(2)^2, or 1 = 4a, or a = 1/4.
The equation of parabola y4 is y+4 = (1/4)x^2
Or you could elim. the fraction and write the eqn as 4y+16=x^2, or
4y = x^2-16, or y = (1/4)x - 4. Take your pick! Hope this helps you find "a" for the other parabolas.
Answer:
c and b
Step-by-step explanation:
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The 1st one
Well it's the same as normal exponents so you multiply 2/3 x 2/3 x 2/3 x 2/3 x 2/3
Answer:
Step-by-step explanation:
<u>Step 1: Make an expression</u>
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<u>Step 2: Cross Multiply</u>
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<u>Step 3: Divide both sides by 4</u>
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Answer:
Answer:
Step-by-step explanation:
So we have the equation:
First, subtract 9 from both sides:
Divide both sides by -4:
Definition of Absolute Value:
Subtract 3 from both sides:
Divide both sides by 5:
And we're done!
