Given function : 3x−6y=12.
We are given x : −2 0 4.
We need to find the values of y's for x=-2, x=0 and x=4.
Plugging x=-2 in the given equation, we get
3(-2)−6y=12
-6 - 6y = 12.
Adding 6 on both sides, we get
-6+6 - 6y = 12+6
-6y = 18.
Dividing by -6 on both sides, we get
y= -3.
On the same way, plugging x=0.
3(0)−6y=12
-6y =12.
y=-2.
Plugging x=4,
3(4)−6y=12
12 -6y = 12.
Subtracting 12 on both sides.
12-12 -6y = 12-12
-6y=0
y=0.
Therefore,
<h3>x −2 0 4</h3><h3>y -3 -2 0</h3>
3x^2 + 21x
Is the answer without a doubt.
<h3>
Answer: 9</h3>
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Work Shown:
a = unknown = leg #1
b = 12 = leg #2
c = 15 = hypotenuse
Plug those values into the pythagorean theorem and solve for 'a'
a^2 + b^2 = c^2
a^2 + (12)^2 = (15)^2 .... substitution
a^2 + 144 = 225
a^2 = 225 - 144 ... subtracting 144 from both sides
a^2 = 81
a = sqrt(81) .... applying square root to both sides
a = 9
F(y) = y + y^2 - 3
f(-2) = -2 + (-2^2) - 3
f(-2) = -2 + 4 - 3
f(-2) = -1
f(-4) = -4 + (-4^2) - 3
f(-4) = -4 + 16 - 3
f(-4) = 9
f(0) = 0 + 0^2 - 3
f(0) = -3
f(2) = 2 + 2^2 - 3
f(2) = 2 + 4 - 3
f(2) = 3
f(4) = 4 + 4^2 - 3
f(4) = 4 + 16 - 3
f(4) = 17
