Number of plants = 5
Amount of soil present for 5 plants = 3/4 gallons
Now here we have to find how much soil will each plant get.
So we have to apply unitary method here to find how much would each plant get the share of soil.
Soil required by 5 plants = 3/4 gallons
Soil required by 1 plant = 3/4 divided by 5 = 3/20 gallons
Answer: Each plant will get 3/20 gallons of soil.
Let the lengths of the sides of the rectangle be x and y. Then A(Area) = xy and 2(x+y)=300. You can use substitution to make one equation that gives A in terms of either x or y instead of both.
2(x+y) = 300
x+y = 150
y = 150-x
A=x(150-x) <--(substitution)
The resulting equation is a quadratic equation that is concave down, so it has an absolute maximum. The x value of this maximum is going to be halfway between the zeroes of the function. The zeroes of the function can be found by setting A equal to 0:
0=x(150-x)
x=0, 150
So halfway between the zeroes is 75. Plug this into the quadratic equation to find the maximum area.
A=75(150-75)
A=75*75
A=5625
So the maximum area that can be enclosed is 5625 square feet.
Parameterize the lateral face

of the cylinder by

where

and

, and parameterize the disks

as


where

and

.
The integral along the surface of the cylinder (with outward/positive orientation) is then




Divide the total by the quantity of numbers:
852/3 = 284
Because there are an odd quantity of numbers (3) the answer is the middle number.
The answer is 284
Answer:
answer is C
Step-by-step explanation:
General equation of a line is expressed as shown:
y = mx+c where;
m is the slope or gradient of the line
c is the intercept of the line
Given the equation of the line graph as y =2.5x
Comparing the given equation with the general equation, it is seen that m = 2.5 and c = 0 (since there is no value for the intercept)
Based on the explanation, the y-intercept of the graph is therefore 0