First of all, we need to know what is the vertex means which is the maximum or minimum point of a parabola and the formula will be:
x=-b/2a
Where b and a from
f(x)=ax^2+bx+c
So do find which function has a vertex of origin. Let's find the vertex of all the function that we had:
f(x)=(x+4)^2
f(x)=(x+4)(x+4)
f(x)=x^2+8x+16
x=-b/2a
x=-8/2(1)
x=-8/2
x=-4
Not the right answer because the vertex needs to be origin which is x=0
f(x)=x(x-4)
f(x)=x^2-4x
x=-b/2a
x=-(-4)/2(2)
x=4/4
x=1
Not the right answer
f(x)=(x-4)(x+4)
f(x)=x^2-16
x=-0/2(1)
x=0
Yay! This is the right answer. As a result, f(x)=(x-4)(x+4) is your final answer. Hope it help!
Answer:
Step-by-step explanation:
we know that
---> by addition angle postulate
we have
----> given problem
----> given problem
substitute in the expression above
Divide by 2 both sides
Mrs. Santos is 29, 44-15=29
Answer:
160m/s
Step-by-step explanation:
The object can hit the ground when t = a; meaning that s(a) = s(t) = 0
So, 0 = -16a² + 400
16a² = 400
a² = 25
a = √25
a = 5 (positive 5 only because that's the only physical solution)
The instantaneous velocity is
v(a) = lim(t->a) [s(t) - s(a)]/[t-a)
Where s(t) = -16t² + 400
and s(a) = -16a² + 400
v(a) = Lim(t->a) [-16t² + 400 + 16a² - 400]/(t-a)
v(a) = Lim(t->a) (-16t² + 16a²)/(t-a)
v(a) = lim (t->a) -16(t² - a²)(t-a)
v(a) = -16lim t->a (t²-a²)(t-a)
v(a) = -16lim t->a (t-a)(t+a)/(t-a)
v(a) = -16lim t->a (t+a)
But a = t
So, we have
v(a) = -16lim t->a 2a
v(a) = -32lim t->a (a)
v(a) = -32 * 5
v(a) = -160
Velocity = 160m/s
Formula for this is as follows:
probability of her passing both 0.6/0.8 - first test and this is a fraction. 0.6/0.8
0.6/0.8= divide 0.6 by 0.8=0.75
that means probability of her passing the second test is 75%
× Sin 50