i need help with my homework! if you know any of these please answer.
1) solve for A if f(x)= -Ax^2+Bx-8, f(-5)= -123 and g(x)=4x+B, g(2)=11
2)Determine the solution set of x^2<2x+8 algebraically
3)The height s of a ball, in feet, thrown with an intial velocity of 112 feet per second from an initial height of 6 ft is given as a function of time t(in seconds) by
s(t)= -16y^2+112t+20.
what is the max height of the ball?
At what time does the max height occur?
1) solve for A if f(x)= -Ax^2+Bx-8, f(-5)= -123 and g(x)=4x+B, g(2)=11
2)Determine the solution set of x^2<2x+8 algebraically
3)The height s of a ball, in feet, thrown with an intial velocity of 112 feet per second from an initial height of 6 ft is given as a function of time t(in seconds) by
s(t)= -16y^2+112t+20.
what is the max height of the ball?
At what time does the max height occur?
-
1) solve for A if f(x)= -Ax² + Bx - 8, f(-5)= -123
-123 = -A(-5)² + B(-5) - 8
-123 = -25A - 5B - 8
-115 = -25A - 5B
-23 = -5A - B
and g(x)=4x+B, g(2)=11
11 = 4(2) + B
11 = 8 + B
3 = B
Plug B into the first equation.
-23 = -5A - 3
-20 = -5A
4 = A
ANSWER: A = 4
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2.) x² < 2x + 8
x² - 2x - 8 < 0
(x - 4)(x + 2) < 0
Answer: http://www.wolframalpha.com/input/?i=pro…
---------------------------------------…
3.) It's a long process, but here's the answer: http://www.wolframalpha.com/input/?i=pro…
:D
-123 = -A(-5)² + B(-5) - 8
-123 = -25A - 5B - 8
-115 = -25A - 5B
-23 = -5A - B
and g(x)=4x+B, g(2)=11
11 = 4(2) + B
11 = 8 + B
3 = B
Plug B into the first equation.
-23 = -5A - 3
-20 = -5A
4 = A
ANSWER: A = 4
---------------------------------------…
2.) x² < 2x + 8
x² - 2x - 8 < 0
(x - 4)(x + 2) < 0
Answer: http://www.wolframalpha.com/input/?i=pro…
---------------------------------------…
3.) It's a long process, but here's the answer: http://www.wolframalpha.com/input/?i=pro…
:D
-
s(t)= -16t²+112t+20.
it is a parabola, open down
its max/min
s' = - 32t+112
s' =0
32t= 112
t= 112/32= 3.5 second
the ball reaches its max in 3.5 seconds
s(3.5)= plug it into the s .
2)Determine the solution set of x^2<2x+8 algebraically
let f(x)=x²-2x-8<0
(x-4)(x+2)<0
x<4 and x> -2
-2
it is a parabola, open down
its max/min
s' = - 32t+112
s' =0
32t= 112
t= 112/32= 3.5 second
the ball reaches its max in 3.5 seconds
s(3.5)= plug it into the s .
2)Determine the solution set of x^2<2x+8 algebraically
let f(x)=x²-2x-8<0
(x-4)(x+2)<0
x<4 and x> -2
-2
1
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