Answer:
5.848 in
Step-by-step explanation:
Divide the piece of cardboard into nine equal squares of side z.
The centre square is the base of the cube.
When you cut out the corner squares and fold up the sides, you will have a cube with edge length z.
The cube must have a volume of 200 in³.
V = z³
200 = z³
![z = \sqrt[3]{200}](https://tex.z-dn.net/?f=z%20%3D%20%5Csqrt%5B3%5D%7B200%7D)
![z = \sqrt[3]{8\times25}](https://tex.z-dn.net/?f=z%20%3D%20%5Csqrt%5B3%5D%7B8%5Ctimes25%7D)
![z = 2\sqrt[3]{25}](https://tex.z-dn.net/?f=z%20%3D%202%5Csqrt%5B3%5D%7B25%7D)
z = 2 × 2.924
z = 5.848 in
The edge length of the cube is 5.848 in.
Answer: Lester gets about 400 k - 700k in casino hiest
Step-by-step explanation:
Answer:
0.293 s
Step-by-step explanation:
Using equations of motion,
y = 66.1 cm = 0.661 m
v = final velocity at maximum height = 0 m/s
g = - 9.8 m/s²
t = ?
u = initial takeoff velocity from the ground = ?
First of, we calculate the initial velocity
v² = u² + 2gy
0² = u² - 2(9.8)(0.661)
u² = 12.9556
u = 3.60 m/s
Then we can calculate the two time periods at which the basketball player reaches ths height that corresponds with the top 10.5 cm of his jump.
The top 10.5 cm of his journey starts from (66.1 - 10.5) = 55.6 cm = 0.556 m
y = 0.556 m
u = 3.60 m/s
g = - 9.8 m/s²
t = ?
y = ut + (1/2)gt²
0.556 = 3.6t - 4.9t²
4.9t² - 3.6t + 0.556 = 0
Solving the quadratic equation
t = 0.514 s or 0.221 s
So, the two time periods that the basketball player reaches the height that corresponds to the top 10.5 cm of his jump are
0.221 s, on his way to maximum height and
0.514 s, on his way back down (counting t = 0 s from when the basketball player leaves the ground).
Time spent in the upper 10.5 cm of the jump = 0.514 - 0.221 = 0.293 s.
Answer:
Number 2
Step-by-step explanation:
Go from the largest negative number then go down the numbers till 0 then go up with the rest of the number is order.
Answer: n>1
Step-by-step explanation:
subtract 1 from both sides
n+1-1>2-1
then simplify
n>1
