Given:
A number is 400.
To find:
The additive inverse of 400.
Solution:
We know that the sum of a number and its additive inverse is 0.
If "a" is number and "b" is its additive inverse, then
Let x be the additive inverse of 400. Then,
Subtract both sides by 400.
Therefore, the additive inverse of 400 is 
.
Answer:
14,850
Step-by-step explanation:
You need the sum of
3 + 6 + 9 + 12 + ... + 294 + 297
Factor out a 3 from the sum
3 + 6 + 9 + 12 + ... + 294 + 297 = 3(1 + 2 + 3 + 4 + ... + 98 + 99)
You need to add all integers from 1 to 99 and multiply by 3.
The sum of all consecutive integers from 1 to n is:
[n(n + 1)]/2
The sum of all consecutive integers from 1 to 99 is
[99(99 + 1)]/2
The sum you need is 3 * [99(99 + 1)]/2
3 + 6 + 9 + 12 + ... + 294 + 297 =
= 3 * [99(99 + 1)]/2
= 3 * [99(100)]/2
= 3 * 9900/2
= 14,850
Answer:
There is no solution
Step-by-step explanation:
-4(x+3)=16-4x
expand
−4x−12=16−4x
cancel out
−12=16
because −12=16 is false there is no solution.
Hope this helps :)
Answer:
cos(O) = 39 / 89
Step-by-step explanation:
Given:
ΔOPQ, where
∠Q=90°
PO = 89
OQ = 39
QP = 80
cosine of ∠O?
cos(O) = Adjacent / Hypotenuse
cos(O) = 39 / 89
