Answer:
The equation you are given is a quadratic. The standard form of a quadratic is y = a(x-h)2 + k where (h,k) is the vertex of the graph, which is a parabola. Vertically moving the graph 4 units upward means that you are moving k +4 units.
y = a(x-h)2 + k standard form
y = 5x2 - 4 original equation
y = 5(x-0)2 - 4 re-written in standard form
h = 0 k = -4
Four (4) units up is k + 4--->-4 + 4 = 0.
Therefore, f(x) = 5x2 + 0--->f(x) = 5x2.
Step-by-step explanation:
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Answer:
x = 12
m(QS) = 52°
m(PD) = 152°
Step-by-step explanation:
Recall: Angle formed by two secants outside a circle = ½(the difference of the intercepted arcs)
Thus:
m<R = ½[m(PD) - m(QS)]
50° = ½[(12x + 8) - (4x + 4)] => substitution
Solve for x
Multiply both sides by 2
2*50 = (12x + 8) - (4x + 4)
100 = (12x + 8) - (4x + 4)
100 = 12x + 8 - 4x - 4 (distributive property)
Add like terms
100 = 8x + 4
100 - 4 = 8x
96 = 8x
96/8 = x
12 = x
x = 12
✔️m(QS) = 4x + 4 = 4(12) + 4 = 52°
✔️m(PD) = 12x + 8 = 12(12) + 8 = 152°
For this case we have the following function transformation:
Vertical displacement.
Assume k> 0
If f (x) is the original function, f (x) + k is the original function with a vertical displacement k units up.
Answer:
Part 1:
C. The function is not shifted horizontally from g (x)
Part 2:
A. 3 units up from g (x)
Answer:
25 minutes
Step-by-step explanation:
he runs 4 laps in 8 minutes
so he runs 1 lap is 8/4 = 2 minutes
Therefore he rub 12.5 laps in 2 * 12.5 = 25 minutes
