![\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{64}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{0\qquad \textit{from the ground}}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Binitial%20velocity%7D%20%5C%5C%5C%5C%20%5Cbegin%7Barray%7D%7Bllll%7D%20~~~~~~%5Ctextit%7Bin%20feet%7D%20%5C%5C%5C%5C%20h%28t%29%20%3D%20-16t%5E2%2Bv_ot%2Bh_o%20%5Cend%7Barray%7D%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Cstackrel%7B64%7D%7B%5Ctextit%7Binitial%20velocity%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h_o%3D%5Cstackrel%7B0%5Cqquad%20%5Ctextit%7Bfrom%20the%20ground%7D%7D%7B%5Ctextit%7Binitial%20height%20of%20the%20object%7D%7D%5C%5C%5C%5C%20h%3D%5Cstackrel%7B%7D%7B%5Ctextit%7Bheight%20of%20the%20object%20at%20%22t%22%20seconds%7D%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)
Check the picture below, it hits the ground at 0 feet, where it came from, the ground, and when it came back down.
56400 ÷ 0.332
<span>169879 total people
</span>
Hope this helps!
Answer:
Step-by-step explanation:
1. 3.8 x 10^5 + 5.5 x 10^5 = 930,000 or 9.3 x 10^5
You just have to solve the equation and shorten the answer. For number one the answer is 930,000, so you have to make it into a number with decimals since a scientific notation number can not be equal to 10 or be greater. So you have to round the number to 9.3 not 93. To "round" it to 9.3 you have to move the decimal at the end of 930,000 to be in between the numbers 9 and 3. How many times you move the decimal to the left or right, that is the exponent of 10. So for 930,000 you moved the decimal 5 times to the left so the exponent for 10 is positive 5.
Answer:
35.34 and 16.66
Step-by-step explanation:
Given data
let the numbers be x and y
the smaller number = x
the larger number = y
x+y=52--------1
<em>The larger number is 2 less than twice the smaller number</em>
y=2x-2-------2
put y= 2x-2 in eqn 1
x+2x-2=52
3x-2=52
3x=52-2
3x=50
x= 50/3
x= 16.66
put x=16.66 in eqn 1 to find y
x+y=52
16.66+y= 52
y=52-16.66
y= 35.34
Hence the numbers are
35.34 and 16.66
Check
35.34+16.66= 52
