ANSWER :
Surface area of larger cone : 24π units^2
Surface area of smaller cone : 6π units^2
The surface area of the smaller cone is 25% of that of the larger cone.
EXPLANATION :
From the given problem,
AB = 3 is the radius of the larger cone and the slanted height is BC = 5
DE = 1.5 is the radius of the smaller cone and the slanted height is EC = 2.5
Recall the surface area of the cone :
![A=\pi r^2+\pi rL](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2%2B%5Cpi%20rL)
where r = radius and L = slanted height.
For the larger cone, r = 3 and L = 5
![\begin{gathered} A=\pi(3)^2+\pi(3)(5) \\ A=24\pi \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A%3D%5Cpi%283%29%5E2%2B%5Cpi%283%29%285%29%20%5C%5C%20A%3D24%5Cpi%20%5Cend%7Bgathered%7D)
For the smaller cone, r = 1.5 and L = 2.5
![\begin{gathered} A=\pi(1.5)^2+\pi(1.5)(2.5) \\ A=6\pi \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A%3D%5Cpi%281.5%29%5E2%2B%5Cpi%281.5%29%282.5%29%20%5C%5C%20A%3D6%5Cpi%20%5Cend%7Bgathered%7D)
Comparing the surface areas :
The area of the smaller cone compared to the larger cone is :
First simplify the expression.
1/3(1,650+0.15c)
550+0.05c
Answer is A
Hope this helps!
Answer:
6 bushes and 2 fences
Step-by-step explanation:
Let 2x be the no. of bushes and 2y be the no. of fences
2x+2y = 8
18(2x)+25(2y) less than/equal to 164
2y = 8 - 2x
18(2x) + 25(8-2x) 《 164
36x + 200 - 50x 《 164
14x 》 36
x 》 18/7
x = 3
y = 1
2x = 6
2y = 2
Answer:
x < -5 or x = 1 or 2 < x < 3 or x > 3
Step-by-step explanation:
Given <u>rational inequality</u>:
![\dfrac{(x-1)^2(x-2)^3}{(x^2-5x+6)^2(x+5)}\geq 0](https://tex.z-dn.net/?f=%5Cdfrac%7B%28x-1%29%5E2%28x-2%29%5E3%7D%7B%28x%5E2-5x%2B6%29%5E2%28x%2B5%29%7D%5Cgeq%200)
![\textsf{Factor }(x^2-5x+6):](https://tex.z-dn.net/?f=%5Ctextsf%7BFactor%20%7D%28x%5E2-5x%2B6%29%3A)
![\implies x^2-2x-3x+6](https://tex.z-dn.net/?f=%5Cimplies%20x%5E2-2x-3x%2B6)
![\implies x(x-2)-3(x-2)](https://tex.z-dn.net/?f=%5Cimplies%20x%28x-2%29-3%28x-2%29)
![\implies (x-3)(x-2)](https://tex.z-dn.net/?f=%5Cimplies%20%28x-3%29%28x-2%29)
Therefore:
![\dfrac{(x-1)^2(x-2)^3}{(x-3)^2(x-2)^2(x+5)}\geq 0](https://tex.z-dn.net/?f=%5Cdfrac%7B%28x-1%29%5E2%28x-2%29%5E3%7D%7B%28x-3%29%5E2%28x-2%29%5E2%28x%2B5%29%7D%5Cgeq%200)
Find the roots by solving f(x) = 0 (set the numerator to zero):
![\implies (x-1)^2(x-2)^3=0](https://tex.z-dn.net/?f=%5Cimplies%20%28x-1%29%5E2%28x-2%29%5E3%3D0)
![\implies (x-1)^2=0\implies x=1](https://tex.z-dn.net/?f=%5Cimplies%20%28x-1%29%5E2%3D0%5Cimplies%20x%3D1)
![\implies (x-2)^3=0 \implies x=2](https://tex.z-dn.net/?f=%5Cimplies%20%28x-2%29%5E3%3D0%20%5Cimplies%20x%3D2)
Find the restrictions by solving f(x) = <em>undefined </em>(set the denominator to zero):
![\implies (x-3)^2(x-2)^2(x+5)=0](https://tex.z-dn.net/?f=%5Cimplies%20%28x-3%29%5E2%28x-2%29%5E2%28x%2B5%29%3D0)
![\implies (x-3)^2=0 \implies x=3](https://tex.z-dn.net/?f=%5Cimplies%20%28x-3%29%5E2%3D0%20%5Cimplies%20x%3D3)
![\implies (x-2)^2=0 \implies x=2](https://tex.z-dn.net/?f=%5Cimplies%20%28x-2%29%5E2%3D0%20%5Cimplies%20x%3D2)
![\implies (x+5)=0 \implies x=-5](https://tex.z-dn.net/?f=%5Cimplies%20%28x%2B5%29%3D0%20%5Cimplies%20x%3D-5)
Create a sign chart, using closed dots for the <u>roots</u> and open dots for the <u>restrictions</u> (see attached).
Choose a test value for each region, including one to the left of all the critical values and one to the right of all the critical values.
Test values: -6, 0, 1.5, 2.5, 4
For each test value, determine if the function is positive or negative:
![f(-6)=\dfrac{(-6-1)^2(-6-2)^3}{(-6-3)^2(-6-2)^2(-6+5)}=\dfrac{(+)(-)}{(+)(+)(-)}=+](https://tex.z-dn.net/?f=f%28-6%29%3D%5Cdfrac%7B%28-6-1%29%5E2%28-6-2%29%5E3%7D%7B%28-6-3%29%5E2%28-6-2%29%5E2%28-6%2B5%29%7D%3D%5Cdfrac%7B%28%2B%29%28-%29%7D%7B%28%2B%29%28%2B%29%28-%29%7D%3D%2B)
![f(0)=\dfrac{(0-1)^2(0-2)^3}{(0-3)^2(0-2)^2(0+5)}=\dfrac{(+)(-)}{(+)(+)(+)}=-](https://tex.z-dn.net/?f=f%280%29%3D%5Cdfrac%7B%280-1%29%5E2%280-2%29%5E3%7D%7B%280-3%29%5E2%280-2%29%5E2%280%2B5%29%7D%3D%5Cdfrac%7B%28%2B%29%28-%29%7D%7B%28%2B%29%28%2B%29%28%2B%29%7D%3D-)
![f(1.5)=\dfrac{(1.5-1)^2(1.5-2)^3}{(1.5-3)^2(1.5-2)^2(1.5+5)}=\dfrac{(+)(-)}{(+)(+)(+)}=-](https://tex.z-dn.net/?f=f%281.5%29%3D%5Cdfrac%7B%281.5-1%29%5E2%281.5-2%29%5E3%7D%7B%281.5-3%29%5E2%281.5-2%29%5E2%281.5%2B5%29%7D%3D%5Cdfrac%7B%28%2B%29%28-%29%7D%7B%28%2B%29%28%2B%29%28%2B%29%7D%3D-)
![f(2.5)=\dfrac{(2.5-1)^2(2.5-2)^3}{(2.5-3)^2(2.5-2)^2(2.5+5)}=\dfrac{(+)(+)}{(+)(+)(+)}=+](https://tex.z-dn.net/?f=f%282.5%29%3D%5Cdfrac%7B%282.5-1%29%5E2%282.5-2%29%5E3%7D%7B%282.5-3%29%5E2%282.5-2%29%5E2%282.5%2B5%29%7D%3D%5Cdfrac%7B%28%2B%29%28%2B%29%7D%7B%28%2B%29%28%2B%29%28%2B%29%7D%3D%2B)
![f(4)=\dfrac{(4-1)^2(4-2)^3}{(4-3)^2(4-2)^2(4+5)}=\dfrac{(+)(+)}{(+)(+)(+)}=+](https://tex.z-dn.net/?f=f%284%29%3D%5Cdfrac%7B%284-1%29%5E2%284-2%29%5E3%7D%7B%284-3%29%5E2%284-2%29%5E2%284%2B5%29%7D%3D%5Cdfrac%7B%28%2B%29%28%2B%29%7D%7B%28%2B%29%28%2B%29%28%2B%29%7D%3D%2B)
Record the results on the sign chart for each region (see attached).
As we need to find the values for which f(x) ≥ 0, shade the appropriate regions (zero or positive) on the sign chart (see attached).
Therefore, the solution set is:
x < -5 or x = 1 or 2 < x < 3 or x > 3
As interval notation:
![(- \infty,-5) \cup x=1 \cup (2,3) \cup(3,\infty)](https://tex.z-dn.net/?f=%28-%20%5Cinfty%2C-5%29%20%5Ccup%20x%3D1%20%5Ccup%20%282%2C3%29%20%5Ccup%283%2C%5Cinfty%29)