A supermarket shop cans display consists of 5 levels of stacked cans. In the display, each level has 4 more cans than the level above it-the total number of cans in display are (b) 100
Step-by-step explanation:
Given that ,
The display consists of 5 levels of stacked cans- <u>which means that there are 5 levels in which the stacked cans are displayed .</u>
so,the no:of levels are 5
Now given that level has 4 more cans than the level above it.
so. the number of cans in a single level = 5*4=20
Now we know that the total number of level is 5 so we multiply 20 withe the number of levels of can stacked
==>20*5=100
<u>Thus we can say that the total number of cans in display are (b) 100</u>
Answer: Roughly 111.97
Step-by-step explanation:
1. 39 divided by 3 is 13.
2. 4367 divided by 3 is not evenly distributed. Rounding, it would work out to be roughly 1455.66
3. 1455.66 divided by 13 is roughly 111.97 (Again, this does not distribute evenly)
4. To check your work, multiply 111.97 times 13 and you should get somewhere around 1455.66 which is (4367 divided by 3).
Hopefully this helps! Feel free to mark brainliest! :)
Answer:
B. A(r(t)) = 25πt²
Step-by-step explanation:
Find the completed question below
The radius of a circular pond is increasing at a constant rate, which can be modeled by the function r(t) = 5t where t is time in months. The area of the pond is modeled by the function A(r) = πr². The area of the pond with respect to time can be modeled by the composition . Which function represents the area with respect to time? A. B. C. D.
Given
A(t) = πr²
r(t) = 5t
We are to evaluate the composite expression A(r(t))
A(r(t)) = A(5t)
To get A(5t), we will replace r in A(t) with 5t and simplify as shown
A(5t) = π(5t)²
A(5t) = π(25t²)
A(5t) = 25πt²
A(r(t)) = 25πt²
Hence the composite expression A(r(t)) is 25πt²
Option B is correct.
Answer:
Step-by-step explanation:
The position function is
and if we are looking for the time(s) that the ball is 10 feet above the surface of the moon, we sub in a 10 for s(t) and solve for t:
and
and factor that however you are currently factoring quadratics in class to get
t = .07 sec and t = 18.45 sec
There are 2 times that the ball passes 10 feet above the surface of the moon, once going up (.07 sec) and then again coming down (18.45 sec).
For part B, we are looking for the time that the ball lands on the surface of the moon. Set the height equal to 0 because the height of something ON the ground is 0:
and factor that to get
t = -.129 sec and t = 18.65 sec
Since time can NEVER be negative, we know that it takes 18.65 seconds after launch for the ball to land on the surface of the moon.
Answer:
Step-by-step explanation:
889
