The answer can be readily calculated using a single variable, x:
Let x = the amount being invested at an annual rate of 10%
Let (8000 - x) = the amount being invested at an annual rate of 12%
The problem is then stated as:
(x * 0.10) + ((8000 - x) * 0.12) = 900
0.10(x) + ((8000 * 0.12) - 0.12(x)) = 900
0.10(x) + 960 - 0.12(x) = 900
0.10(x) - 0.12(x) = 900 - 960
-0.02(x) = -60
-0.02(x) * -100/2 = -60 * -100/2
x = 6000 / 2
x = 3000
Thus, $3,000 is invested at 10% = $300 annually; and $8,000 - $3,000 = $5,000 invested at 12% = $600 annually, which sum to $900 annual investment.
Answer:
270 dollars for a year, so 22.50 times 12
Depends on the speed of the road assuming 65 or faster no they were not speeding. At 60 yes they were but not by much lol
The answer is 2x(2x²+x+1).
When we subtract polynomials we combine like terms:
(9x³+2x²-5x+4)-(5x³-7x+4)
9x³-5x³=4x³
2x²- 0 = 2x²
-5x--7x=-5x+7x=2x
4-4=0
This gives us
4x³+2x²+2x
Each of these is divisible by 2, and each has an x, so we factor those out:
2x( )
4x³/2x = 2x²:
2x(2x² )
2x²/2x=x:
2x(2x²+x )
2x/2x = 1:
2x(2x²+x+1)
(2, 12)
Because that is where the graphs intersect.
