420/100= 42x10/10x10=
42/10=
2x21/2x5=
21/5
Answer:
Yes, it is.
Step-by-step explanation:
(4) = -(4)+8
4=4
True
OB. h(Z) = 412 is the answer
Answer:
The capacity of the container is 2546.78 cm³.
Step-by-step explanation:
The volume of the frustum of a cone is:
![\text{Volume}=\frac{\pi h}{3}\cdot[R^{2}+Rr+r^{2}]](https://tex.z-dn.net/?f=%5Ctext%7BVolume%7D%3D%5Cfrac%7B%5Cpi%20h%7D%7B3%7D%5Ccdot%5BR%5E%7B2%7D%2BRr%2Br%5E%7B2%7D%5D)
The information provided is:
r = 16/2 = 8 cm
R = 24/2 = 12 cm
h = 8 cm
Compute the capacity of the container as follows:
![\text{Volume}=\frac{\pi h}{3}\cdot[R^{2}+Rr+r^{2}]](https://tex.z-dn.net/?f=%5Ctext%7BVolume%7D%3D%5Cfrac%7B%5Cpi%20h%7D%7B3%7D%5Ccdot%5BR%5E%7B2%7D%2BRr%2Br%5E%7B2%7D%5D)
![=\frac{\pi\cdot8}{3}\cdot[(12)^{2}+(12\cdot 8)+(8)^{2}]\\\\=\frac{8\pi}{3}\times [144+96+64]\\\\=\frac{8\pi}{3}\times304\\\\=2546.784445\\\\\approx 2546.78](https://tex.z-dn.net/?f=%3D%5Cfrac%7B%5Cpi%5Ccdot8%7D%7B3%7D%5Ccdot%5B%2812%29%5E%7B2%7D%2B%2812%5Ccdot%208%29%2B%288%29%5E%7B2%7D%5D%5C%5C%5C%5C%3D%5Cfrac%7B8%5Cpi%7D%7B3%7D%5Ctimes%20%5B144%2B96%2B64%5D%5C%5C%5C%5C%3D%5Cfrac%7B8%5Cpi%7D%7B3%7D%5Ctimes304%5C%5C%5C%5C%3D2546.784445%5C%5C%5C%5C%5Capprox%202546.78)
Thus, the capacity of the container is 2546.78 cm³.
X pounds of $3
y pounds of $3.25
x+y = 200
300x + 325y = 320*200
300x + 325y = 64000
x+y =200
300x +325y =64000
Solve by eliminating
Multiply top one w/-300
-300y =60000
325y=64000
25y = 4000
y= 160 then
x + 160= 200
x = 40
So the mixture has;
40 pounds of $3 coffee
160 pounds of $3.25 coffee