Answer:
Step-by-step explanation:
h = -6
+20t + 4
I will use calculus, maybe that's not how you're supposed to do this
-12t +20 =0
12t = 20
t = 20 /12
t = 1 
t = 1 
there will be a max at 1.6666666666 seconds
-6*
+ 20 * 1.6666666666 +4
= 16.666666666666 + 33.333333333333 + 4
= - 16 + 33 
+ 4
= 20 feet max height ( not too high, for a rocket)
time of flight:
0 = -6 +20t + 4
use quadratic formula to find t
-20 +- sqrt [ 
- 4*(-6)*4 ] / 2*(-6)
-20 +- sqrt [400 + 96 ] / -12
-20 +- sqrt [496 ] / -12
-20 +- 22.27105 / -12
try the negative option 1st
-42.27105 / -12
3.522 seconds. time of flight
when will the rocket be at 12' ? :
12 = -6 +20t + 4
0 = -6 +20t -8
use quadratic formula again to find t
-20 +- sqrt [ - 4*(-6)*(-8) ] / 2*(-6)
-20 +- sqrt [ 400 - 192 ] / -12
-20 +- sqrt [208 ] / -12
-20 - 14.4222 / -12
-34.4222 / -12
2.8685 seconds ( on the way down)
and
-20 + 14.4222 / -12
-5.578 / - 12
0.4648 seconds ( on the way up )
Answer:
La arista de la bodega debe medir 6 metros.
Step-by-step explanation:
Podemos encontrar la arista de la bodega sabiendo su volumen:
En donde:
V: es el volumen de un cubo = 216 m³
a: es la arista del cubo =?
La arista del cubo es:
![a = \sqrt[3]{V} = \sqrt[3]{216 m^{3}} = 6 m](https://tex.z-dn.net/?f=%20a%20%3D%20%5Csqrt%5B3%5D%7BV%7D%20%3D%20%5Csqrt%5B3%5D%7B216%20m%5E%7B3%7D%7D%20%3D%206%20m%20)
Por lo tanto, la arista de la bodega debe medir 6 metros.
Espero que te sea de utilidad!
The answer to your problem would be 10.
if two figures are congruent, each side is congruent to the same side on the other figure.
AB=FG; CD=HI; EA=JF
since CD=HI, and CD=10, then HI would also equal 10
434/7 = 62/1
Hope this helped in any way :)
Minus 150 meters pepega I don’t quite understand the question but that is one way to get from 400 to 350 meters.
