700/32 = 21.875
He'd have to sell approximately 22 shares to have enough to pay for his vacation.
Answer:
-3
The problem:
The sum of four consecutive integers is - 18. What is the greatest of these
integers?
Step-by-step explanation:
If n is the first integer, then n+1 is the second integer, n+2 is the third integer, and n+3 is the fourth integer in consecutive order.
For example if n=8, we are saying n+1=9,n+2=10, and n+3=11, which I think you can see that 8,9,10,11 are consecutive.
So the sum is -18 which means:
n+(n+1)+(n+2)+(n+3)=-18
4n+6=-18
Subtract 6 on both sides:
4n=-18-6
Simplify:
4n=-24
Divide both sides by 4:
n=-6
If n=-6,
then:
n+1=-6+1=-5
n+2=-6+2=-4
n+3=-6+3=-3.
So the 4 consecutive integers whose sum is -18 is: -6,-5,-4, and -3.
The greatest of these integers is -3.
Answer:
That should be 11.99+35.50+6.75+3+ 7.15% of EACH of those items added together minus 11.99+35.50+6.75+3+ 6.35% of EACH of the same items added together
Answer:
(7x+5)(7x-5)
Step-by-step explanation:
We can use the difference of squares equation which states that when a number is multiplied like: (ax+b)(ax-b) then it equals (ax-b). In this equation, since there is no "b" in the typical ax^2+bx+c, we can recognize that the answer is not a square, but a difference of squares.
-13 is lower than -10, therefore the question would be -13<-10
