Answer: 5 cookies
Step-by-step explanation: 8 + .4c = 2.5 + 1.5c
8 + 0.4c = 2.5 + 1.5c
8 · 10 + 0.4c · 10 = 2.5 · 10 + 1.5c · 10
80 + 4c = 25 + 15c
4c = 15c - 55
4c - 15c = 15c - 55 - 15c
-11c = -55
c = 5
The sum of the first n odd numbers is n squared! So, the short answer is that the sum of the first 70 odd numbers is 70 squared, i.e. 4900.
Allow me to prove the result: odd numbers come in the form 2n-1, because 2n is always even, and the number immediately before an even number is always odd.
So, if we sum the first N odd numbers, we have
The first sum is the sum of all integers from 1 to N, which is N(N+1)/2. We want twice this sum, so we have
The second sum is simply the sum of N ones:
So, the final result is
which ends the proof.
Answer:
The square root of 121.
Step-by-step explanation:
The square root of 121 is 11. 11x11=121 which makes the square root of 121, 11. The cube root of 125 is 5. Cube roots are numbers that are multiplied by each other to make a number, 125 is 5x5x5 so the cube root is 5. 11 is a greater number than 5 which makes the square root of 121 greater.
Answer:
90, 91 and 92
Step-by-step explanation:
Given
Consecutive integers = 273
Required
Find the integers
The question seem to be incomplete. However, I'll assume we're dealing with sum.
Let the smallest integer be y.
So,
y + y + 1 + y + 2 = 273
Collect like terms.
y + y + y = 273 - 2 - 1
3y = 270
Divide both sides by 3
y = 90
Hence, the integers are 90, 91 and 92
