Total pay before deductions is gross pay <==
total pay after deductions is net pay
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:
-3
Step-by-step explanation:
y = -3x
This is in slope intercept form
y = mx+b where m is the slope and b is the y intercept
The slope is -3 and the y intercept is 0
Choice B would be the correct answer. I hope that helps! :D
We use the Work formula to solve for the unknown in the problem which is W = F x d. First, we solve for the Net Force acting on the car. The Net Force is the summation of all forces acting on the object. For this case, we assume that Friction Force is negligible thus the Net Force is equal to:
F = mgsinα in terms of SI units and in terms of english units we have F = m(g/g₀)(sin α) where g₀ is the proportionality factor, 32.174 ft lb-m / lb-f s²
F = 2500 (32.174/32.174) (sin 12°) = 519.78 lb
W = Fd = 519.78 lb (400 ft) = 207912 ft - lb or 20800 ft-lb
