Answer:
A
Step-by-step explanation:
To find how long it takes for the object to hit the ground you need to set h(t)=0 after this you need to find the number that when plugged into t makes the equation equal to 0
This is what i got in the caculator 6400 if not im so so sory
Answer: 983.33
Step-by-step explanation:
(0.5+0.5) x 983.33
1 x 983.33 = 983.33
Answer:
a
Step-by-step explanation:
![\sqrt[4]{144a^{12}b^{3}} = \sqrt[4]{4^{2}*3^{2}a^{12}b^{3}}=\\= \sqrt[4]{2^{4}*3^{2}a^{12}b^{3}}=2a^{3}\sqrt[4]{3^{2}b^{3}} =\\}=2a^{3}\sqrt[4]{9b^{3}}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B144a%5E%7B12%7Db%5E%7B3%7D%7D%20%3D%20%5Csqrt%5B4%5D%7B4%5E%7B2%7D%2A3%5E%7B2%7Da%5E%7B12%7Db%5E%7B3%7D%7D%3D%5C%5C%3D%20%5Csqrt%5B4%5D%7B2%5E%7B4%7D%2A3%5E%7B2%7Da%5E%7B12%7Db%5E%7B3%7D%7D%3D2a%5E%7B3%7D%5Csqrt%5B4%5D%7B3%5E%7B2%7Db%5E%7B3%7D%7D%20%3D%5C%5C%7D%3D2a%5E%7B3%7D%5Csqrt%5B4%5D%7B9b%5E%7B3%7D%7D)
