Answer:
Opps i am really sorry i donot know it
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.
Answer:
B.
Step-by-step explanation:
I think your question is missed of key information, allow me to add in and hope it will fit the original one.
Please have a look at the attached photo.
My answer:
As given in the question, we know that:
The ratio of the area of the circle to the area of the square is π/4
- The formula to find the volume of the cone is:
V = 1/3*the height*the base area
<=> V1 = 1/3*h*π
- The formula to find the volume of the pyramid is:
V2 = 1/3*the height*the base area
<=> V = 1/3*h*4
=> the ratio of volume of the cone to the pyramid is:
=
= (1/3*h*π ) / ( 1/3*h*4 )
= π/4
S we can conclude that the volume of the cone equals π/4 the volume of the pyramid
Hope it will find you well.
Answer:
T'
Step-by-step explanation:
See the diagram attached.
This is a unit circle having a radius (r) = 1 unit.
So, the length of the circumference of the circle will be 2πr = 2π units.
Now, the point on the circle at a distance of x along the arc from P is T.
Therefore, the point on the circle at a distance of 2π - x along the arc from P will be T' , where, T' is the image of point T, when reflected over the x-axis. (Answer)