Answer:
(1/5)·100
Step-by-step explanation:
The number of jasmine flowers is 1/5 of the total number of flowers, so an expression for the number that are jasmine is ...
1/5×100
Answer:
![f(x) = 2x^2-3](https://tex.z-dn.net/?f=f%28x%29%20%3D%202x%5E2-3)
Step-by-step explanation:
You are missing information, but the question gives enough information for me to figure it out.
That function shifted down three units is :
. If your answer choice is a function, choose this one. If it is a graph, please use a graphing calculator, desmos, or another graphing website to compare it with.
That's a tough one maybe he's gone to get milk Orr he's probably lost on Antarctica
Let width = w
Let length = l
Let area = A
3w+2l=1200
2l=1200-3w
l=1200-3/2
A=w*l
A=w*(1200-3w)/2
A=600w-(3/2)*w^2
If I set A=0 to find the roots, the maximum will be at wmax=-b/2a which is exactly 1/2 way between the roots-(3/2)*w^2+600w=0
-b=-600
2a=-3
-b/2a=-600/-3
-600/-3=200
w=200
And, since 3w+2l=1200
3*200+2l=1200
2l = 600
l = 300
The dimensions of the largest enclosure willbe when width = 200 ft and length = 300 ft
check answer:
3w+2l=1200
3*200+2*300=1200
600+600=1200
1200=1200
and A=w*l
A=200*300
A=60000 ft2
To see if this is max area change w and l slightly but still make 3w+2l=1200 true, like
w=200.1
l=299.85
A=299.85*200.1
A=59999.985
The distance covered at time t is listed below:
Distance at 2 sec = d(2) = 64 feet
Distance at 4 sec = d(4) = 256 feet
Distance at 6 sec = d(6) = 576 feet
Distance at 8 sec = d(8) = 1024 feet
We are to find the average rate of change between 2 seconds and 6 seconds. The average rate of change will be:
![\frac{d(6)-d(2)}{6-2} \\ \\ = \frac{576-64}{4} \\ \\=128](https://tex.z-dn.net/?f=%20%5Cfrac%7Bd%286%29-d%282%29%7D%7B6-2%7D%20%5C%5C%20%5C%5C%20%3D%20%5Cfrac%7B576-64%7D%7B4%7D%20%5C%5C%20%5C%5C%3D128)
Therefore, the average rate of distance between 2 seconds and 6 seconds is 128 feet per second. This represents the speed of the object. So the speed of the object between 2 and 6 seconds was 128 feet per second.