Step-by-step explanation:
after I answered the other one(s), you should be really able to do this yourself.
it is precisely the same method and formula just with different numbers.
what's the problem now ?
y = -16x² + 247x + 141
we assume, the ground is at 0 ft.
so, we need to solve
0 = -16x² + 247 x + 141
the general solution for such quadratic equations is
x = (-b ± sqrt(b² - 4ac))/(2a)
in our case
a = -16
b = 247
c = 141
x = (-247 ± sqrt(247² - 4×-16×141))/(2×-16) =
= (-247 ± sqrt(61009 + 9024))/-32 =
= (-247 ± sqrt(70033))/-32
x1 = (-247 + 264.6374879...)/-32 = -0.551171497... s
x2 = (-247 - 264.6374879...)/-32 = 15.9886715... s
again, the negative solution for time did not make any sense, so, x2 is our solution.
the rocket will hit the ground after about 15.99 seconds.
This is an altitude. Shown by the right angle on the side opposite the point.
Answer:
y+2=1/2(x-8) or y+4=1/2(x-4)
Step-by-step explanation:
Okay, so point slope form is y-y1=m(x-x1). So you substitute y1 with -2 or 8 and x1 with 8 or 4.
Slope is (-2-(-4))/(8-4), which is 2/4, or 1/2. Plug that into m.
Answer:
No
Step-by-step explanation:
the side lengths would need to add up to 180, these add up to 179
Answer:
b = 12
Step-by-step explanation:
c^2 = a^2 + b^2
Solving for b,
b^2 = c^2 - a^2
= (13)^2 - (5)^2
= 169 - 25
= 144
or
b = 12
