<span>I am assuming that this is a parametric curve.
We see that the curve intersects the x-axis when:
t - t^2 = 0 ==> t = 0 and t = 1.
Then, since x = 1 + e^t is an increasing function, the curve is being traced exactly once on the interval (0, 1).
Using the fact that the area under the curve given by the parametric equations x = f(t) and y = g(t) on (a, b) is:
A = ∫ f'(t)g(t) dt (from t=a to b),
and that f(t) = 1 + e^t ==> f'(t) = e^t, the area under the curve is:
A = ∫ e^t(t - t^2) dt (from t=0 to 1)
= e^t(-t^2 + 3t - 3) (evaluated from t=0 to 1), by integrating by parts
= e(-1 + 3 - 3) - (0 + 0 - 3)
= 3 - e. </span>
77 is the answer because 3×7=21 21+56=77
Continuous I think :/ I am pretty sure like 93% sure
Answer:
I believe the correct answer is B.
Answer:
y≤6x+8
Step-by-step explanation:
<u><em>Given function:</em></u>
<u><em>This means:</em></u>
variable stays same
if y=7 then x=1 and if y=14 then x can be 2 or 3.
<u><em>we can also think as:</em></u>
y≤6x+8
<u><em>what we did:</em></u>
double the number: 3x+4 · 2 = 6x+8
Therefore, another equation for y≤3x+4 would be y≤6x+8.