300 chocolate chips is 6 batches of 50 chips each
(300 chocolate chips = 6 batches @ 50 chips ea)
Each batch of 50 chips = 1.1 cups of flour
Therefore 6 batches @ 1.1 cups ea = 6.6 cups of flour
-----------------
0.6 cups (3/5 cups) of flour makes 6 cookies
6.6 cups of flour is 11 times more than that.
(0.6 x 11 = 6.6)
So she can make 11 times as many cookies.
(11 x 6 cookies = 66 cookies)
Answer:
Bella can make 66 cookies
(minus the three cookies that SumDude ate)
The answer is 0 because any number elevated to 0 is 0
Answer:
A≈113.1
Step-by-step explanation:
A=πr2
A=π·62
A≈113.09734
Answer:
The statement is true is for any 
.
Step-by-step explanation:
First, we check the identity for 
:
The statement is true for .
Then, we have to check that identity is true for 
, under the assumption that 
is true:
![(1^{2}+2^{2}+3^{2}+...+k^{2}) + [2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}](https://tex.z-dn.net/?f=%281%5E%7B2%7D%2B2%5E%7B2%7D%2B3%5E%7B2%7D%2B...%2Bk%5E%7B2%7D%29%20%2B%20%5B2%5Ccdot%20%28k%2B1%29-1%5D%5E%7B2%7D%20%3D%20%5Cfrac%7B%28k%2B1%29%5Ccdot%20%5B2%5Ccdot%20%28k%2B1%29-1%5D%5Ccdot%20%5B2%5Ccdot%20%28k%2B1%29%2B1%5D%7D%7B3%7D)
![\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)}{3} +[2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}](https://tex.z-dn.net/?f=%5Cfrac%7Bk%5Ccdot%20%282%5Ccdot%20k%20-1%29%5Ccdot%20%282%5Ccdot%20k%20%2B1%29%7D%7B3%7D%20%2B%5B2%5Ccdot%20%28k%2B1%29-1%5D%5E%7B2%7D%20%3D%20%5Cfrac%7B%28k%2B1%29%5Ccdot%20%5B2%5Ccdot%20%28k%2B1%29-1%5D%5Ccdot%20%5B2%5Ccdot%20%28k%2B1%29%2B1%5D%7D%7B3%7D)
![\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot [2\cdot (k+1)-1]^{2}}{3} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}](https://tex.z-dn.net/?f=%5Cfrac%7Bk%5Ccdot%20%282%5Ccdot%20k%20-1%29%5Ccdot%20%282%5Ccdot%20k%20%2B1%29%2B3%5Ccdot%20%5B2%5Ccdot%20%28k%2B1%29-1%5D%5E%7B2%7D%7D%7B3%7D%20%3D%20%5Cfrac%7B%28k%2B1%29%5Ccdot%20%5B2%5Ccdot%20%28k%2B1%29-1%5D%5Ccdot%20%5B2%5Ccdot%20%28k%2B1%29%2B1%5D%7D%7B3%7D)
![(2\cdot k +1)\cdot [k\cdot (2\cdot k -1)+3\cdot (2\cdot k +1)] = (k+1) \cdot (2\cdot k +1)\cdot (2\cdot k +3)](https://tex.z-dn.net/?f=%282%5Ccdot%20k%20%2B1%29%5Ccdot%20%5Bk%5Ccdot%20%282%5Ccdot%20k%20-1%29%2B3%5Ccdot%20%282%5Ccdot%20k%20%2B1%29%5D%20%3D%20%28k%2B1%29%20%5Ccdot%20%282%5Ccdot%20k%20%2B1%29%5Ccdot%20%282%5Ccdot%20k%20%2B3%29)
Therefore, the statement is true for any .
Answer:
the answer is the second one
