.27 repeating.... you have to put three on the inside and 11 outside, so 11 goes into 3 zero times so put a zero at the top. Write a zero underneath and subtract 3-0 (obviously 3) and put a decimal point after the first three under the sign and the 0 above it. Add a zero after the threes decimal, and carry it down to the bottom most 3 to have 30. 11 goes into 30 two times so write a 2 at the top. Subtract the 30-22 at the bottom and get 8. Add another 0 after the 3 ( so now 3.00) and carry it to the 8 (80). 11 goes in 7 times, 77. So 7 on top and 80-77= 3 below. Add another zero to repeat process if necessary, but otherwise it repeats, just so you know ;) hope this helps
Answer:
t=6
Step-by-step explanation:
ground height = 0
(are you sure your formula is correct? isn't it - 16t²?)
if h=16t² +64t+192 is true then
16t² +64t+192 = 0
t² + 4t + 12 = 0
t = (-4 ± √(4² - 4*12)) / 2*1 = (-4 ± √-32) / 2 = -2 ± 2√-2
There is no solution of t
if it is h= - 16t² +64t+192
0 =- 16t² + 64t + 192
t² - 4t - 12 = 0
(t + 2) (t -6) = 0
t should be positive
t = 6 sec
Hello!
This is a problem about the general solution of a differential equation.
What we can first do here is separate the variables so that we have the same variable for each side (ex. 
with the 
term and 
with the 
term).
Then, we can integrate using the power rule to get rid of the differentiating terms, remember to add the constant of integration, C, to at least one side of the resulting equation.
Then here, we just solve for and we have our general solution.
![y=\sqrt[3]{\frac{1}{2}x^2-x+C}](https://tex.z-dn.net/?f=y%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B1%7D%7B2%7Dx%5E2-x%2BC%7D)
We can see that answer choice D has an equivalent equation, so answer choice D is the correct answer.
Hope this helps!
Answer:
Step-by-step explanation:
<u>Use identities to set the equation up as a quadratic</u>
<u>Make the substitution u=cos(x) and solve the quadratic</u>
<u />
Hence,
