Question

Few questions

Answer 1

11. If an object is dropped from a height of 200 feet, the function h(t)=-16t^2=200 gives the height of the object after t seconds. Approximately, when will the object hit the ground? (1point)
A.200.00 seconds
B.184.00 seconds
C.3.54 seconds
D.0.78 seconds

Answer 2

13. A catapult launches a boulder with an up ward velocity of 122 feet per second. The height of the boulder,(h), in feet after t seconds is given by the function h(t)=-16t^2+122t+10. How long does it take the boulder to reach its maximum height? Round to the nearest hundredth, if necessary.

Answer 3

11. solve for t when h(t) = 0

-16t^2 + 200 = 0

200 = 16t^2

solve for t.


13. maximum height is when h'(t) = 0 [first derivative of h(t)]

h'(t) = -32t + 122

when h'(t) = 0

32t = 122

solve for t.

Answer 4

thanks i got them!!!

20. solve the equation by completing the square. round to the nearest hundreth.
x^2+6x=-7

A.4.41,1.59
B.-4.41,1.59
c.-4.41,-1.59
D.4.41,-1.59

@Euler271

Answer 5

21.solve the equation by completeing the square.
x^2-6x+7=0

A.-1.59,4.41
B.-1.59,-4.41
C.1.59,-4.41
D.1.59,4.41

@Euler271

Answer 6

22. Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth.

x^2-12=-4x

A.6,2
B.6,-2
C.-6,2
D.-6,-2

@Euler271

Answer 7

i have to go to bed but ill show some steps.

20.

x^2 + 6x + 7 = 0 = (x+3)^2 - 9 + 7

(x+3)^2 = 2

x+3 = ±sqrt(2)

x = ±sqrt(2) - 3

so C


21 is close:

x^2 - 6x + 7 = 0 = (x-3)^2 - 9 + 7

(x-3)^2 = 2

x-3 = ±sqrt(2)

x = ±sqrt(2) + 3

so D

22. use quadratic equation.

Answer 8

could you do 22 real quick please?? @Euler271

Answer 9

but thank you so much

Answer 10

quadratic:
\[x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2 }\]
on your polynomial ax^2 + bx +c = 0

\[x = \frac{ -4 \pm \sqrt{4^2-4(1)(-12)} }{ 2(1) } = \frac{ -4 \pm \sqrt{ 64}}{ 2 } = -2 \pm 4\]

answer is C

Answer 11

wait i don't get 11! how did u solve for t from 200=16^2 ????