Answer:
The vertex of this parabola, 
, can be found by completing the square.
Step-by-step explanation:
The goal is to express this parabola in its vertex form:
,
where 
, 
, and 
are constants. Once these three constants were found, it can be concluded that the vertex of this parabola is at 
.
The vertex form can be expanded to obtain:
.
Compare that expression with the given equation of this parabola. The constant term, the coefficient for 
, and the coefficient for 
should all match accordingly. That is:
.
The first equation implies that is equal to 
. Hence, replace the "
" in the second equation with 
to eliminate 
:
.
.
Similarly, replace the "" and the "" in the third equation with and 
, respectively:
.
.
Therefore, 
would be equivalent to 
. The vertex of this parabola would thus be:
.
<h3>Answer: C) none of the equations are identities</h3>
If you plugged theta = 0 into the first equation, then you would have
sin(45) + cos(45) = sin(0) + cos(0)
sqrt(2) = 1
which is a false equation. We don't have an identity here.
The same story happens with the second equation. Plug in theta = 0 and it becomes
cos(60) - sin(60) = cos^2(0) + tan(0)
1/2 - sqrt(3)/2 = 1 + 0
-0.37 = 1
which is false.
Plug in x-values and see which one has an incorrect y value.
(x, y)
x=0; y=0+7=7; CORRECT
x=2; y=2+7=9; INCORRECT
x=9; y=9+7=16; CORRECT
x=12; y=12+7=19; CORRECT
The point that is not on the online is (9, 16).
Answer:
B
Step-by-step explanation:
In the composition of <em>f</em><em> </em>and <em>g</em>, the equation is telling us to place the function <em>f</em> inside <em>g</em>. Observing in B that <em>f</em> is (x + 4), then this can be put in place of the x in 3x^2, making it 3(x + 4)^2 and finally the minus 5.
