Answer:
YOU not smartYOU DONT KNOW THIS HA HA NOT HELPING YOU BOT.
Step-by-step explanation:
Answer:
The surface area of Design A is smaller than the surface area of Design B.
The area of Design A is 94.65% of the Design B.
Step-by-step explanation:
The area of a right cone is given by the sum of the circle area of the base and the lateral area:
(1)
Where:
r: is the radius
L: is the slant height
The slant height is related to the height and to the radius by Pitagoras:
(2)
By entering equation (2) into (1) we have:
Now, let's find the area of the two cases.
Design A: height that is double the diameter of the base, H= 2D = 4r
The volume of the cone is:

We can find "r":

The area is:
Design B: height that is triple the diameter of the base, H = 3D = 6r
The radius is:
The area is:
Hence, the surface area of Design A is smaller than the surface area of Design B.
The percent of the surface area of Design A is less than Design B by:

Therefore, the area of Design A is 94.65% of the Design B.
I hope it helps you!
There is no picture here that or you need to be more pacific
The price of one senior citizen ticket is 8$ and one student ticket is 12$.
<h3>What is the equation?</h3>
The equation is defined as mathematical statements that have a minimum of two terms containing variables or numbers that are equal.
Let the price of one senior citizen ticket = x
And the price of one student ticket = y
Given that, on the first day of ticket sales, the school sold 13 senior citizen tickets and 13 student tickets for a total of $260
The school took in $212 on the second day by selling 13 senior citizen tickets and 9 student tickets.
13x +13y = 260
13x + 9y = 212
Subtract the equation from first
13x +13y - (13x + 9y) = 260 - 212
4y = 48
y = 48/4
y = 12
Substitute the value of y in the equation 13x + 9y = 212
13x + 9(12) = 212
13x + 108 = 212
13x = 212 - 108
13x = 104
x = 104/13
x = 8
Hence, the price of one senior citizen ticket is 8$ and one student ticket is 12$.
Learn more about the equation here:
brainly.com/question/10413253
#SPJ1