Answer:
It is given that the volume of a cone = 
cubic units
Volume of cone with radius 'r' and height 'h' = 
Equating the given volumes, we get
× 
It is given that the height is 'x' units.
Therefore, 
Therefore, 
So, the expression '
' represents the radius of the cone's base in units.
![\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.
Answer:
50
Step-by-step explanation:
Use pythagoreom theorem to determine the length of the hypotenuse:
a² + b² = c²
- assign side a to 30 and side b to 40
30² + 40² = c²
90 + 160 = c²
250 = c²
Now square root both sides to isolate side c:
√250 = √c²
50 = c
side c = 50
Answer:
infinite many
Step-by-step explanation:
If they're asking for "congruent" which we know is the shape is the same or equal to. Out of the transformations I assume Dilation because you're altering the size of the shape meaning the image is either bigger than or less than the pre-image. Of course a shape bigger is not equal to the original shape.
think of a king size candy bar to a fun size bar. they're not congruent or the same
