Answer:
C. 0, -1, 3
Step-by-step explanation:
Note: There is no graph. I assume you mean "zeroes/roots" by "solutions".
x(x^2-2x-3)=0
/(x^2-2x-3)
x=0
x^2-2x-3 = x^2+x-3x-3 = x(x+1) + ( -3(x+1) = (x-3)(x+1)
/(x-3) --> x +1 =0, x=-1
/(x+1) --> x-3 = 0, x=3
Answer:
1.2
Step-by-step explanation:
You simply have to look for a point on the graph that has an x-coordinate of -6. Only one is (-6. 1.2), so 1.2 is your answer.
9/16+ 1/2
= 9/16+ (1*8) / (2*8)
= 9/16+ 8/16 (common denominator is 16)
= (9+8)/16
= 17/16
= (16+1)/16
= 16/16+ 1/16
= 1+ 1/16
= 1 1/16
The final answer is 1 1/16~
Answer:
2-1=1
1+50=51
Step-by-step explanation:
Answer:
3
Step-by-step explanation:
lim(t→∞) [t ln(1 + 3/t) ]
If we evaluate the limit, we get:
∞ ln(1 + 3/∞)
∞ ln(1 + 0)
∞ 0
This is undetermined. To apply L'Hopital's rule, we need to rewrite this so the limit evaluates to ∞/∞ or 0/0.
lim(t→∞) [t ln(1 + 3/t) ]
lim(t→∞) [ln(1 + 3/t) / (1/t)]
This evaluates to 0/0. We can simplify a little with u substitution:
lim(u→0) [ln(1 + 3u) / u]
Applying L'Hopital's rule:
lim(u→0) [1/(1 + 3u) × 3 / 1]
lim(u→0) [3 / (1 + 3u)]
3 / (1 + 0)
3