A model rocket is launched from a platform 12 meters high at a speed of 35 meters per second. Its height h can be modeled by the equation h= -4.9t'2+35t+12 where t is the time in seconds. at what time will the rocket be altitude of 60 meters.

Step-by-step explanation:

so set the given equation equal to 60 and solve for t

60= -4.9 +35t +12

0= -4.9 +35t - 48

now use the quadratic equation to solve for t

-b +- sq rt ( - 4 a c )  /  2 a

t = -35 +- sq. rt [ 1225 -4(-4.9)(-48) ] / 2 (-4.9)

t = -35 +- sq. rt [ 284.2] / -9.8

t = -35 +- 16/858 / - 9.8

t = 3.571 +- 1.72

t  = 1.85 seconds ( on the way up ) or  t = 5.29 seconds ( on the way down)

at both of those times the rocket will be at 60 meters  :)

13

step-by-step explanation:

169-1=168

168-3=165

165-5=160

160-7=153

153-9=144

144-11=133

133-13=120

120-15=105

105-17=88

88-19=69

69-21=48

48-23=25

25-25=0

since it took 13times to completely subtract it hence its square root is 13

answer: 16

step-by-step explanation: first you would add 6+6 which equals 12 then add 2+2 which equals 4. then add 12 + 4 which is 16.

hope this ^_^