Answer:
490
Step-by-step:
90•5
The absolute value measures the distance a number is away from the origin (zero) on the number line. A number can be to the left or right of zero on the number line, but the distance away from zero is always positive. Graphically:
In other words:
• If d is positive and

• If d is negative and

Therefore, the equation that has no solutions is:
D) f(n) = 3 + 4(n-1)
3 = 1st term
4 = common difference among the terms
n = term number you are looking for.
To check: 3, 7, 11, 15, ...
f(1) = 3 + 4(1-1) = 3 + 4(0) = 3 + 0 = 3
f(2) = 3 + 4(2-1) = 3 + 4(1) = 3 + 4 = 7
f(3) = 3 + 4(3-1) = 3 + 4(2) = 3 + 8 = 11
f(4) = 3 + 4(4-1) = 3 + 4(3) = 3 + 12 = 15
3(x^2+10x+5)-5(x-k)=
3x^2+30x+15-5x+5k=
3x^2+25x+15+5k
for this to be divisible by x every term must include x or get eliminated
the problematic terms are 15 and 5k
to eliminate them they must equal 0 when added:
15+5k=0
5k=-15
k=-3
so A) -3 is the solution