A because the line is only going through the graph and hitting each point once
I think the correct answer from the choices listed above is option C. <span>When you make the circle bigger or smaller, it is the radius of the standard equation for the circle centered at the origin changes. Hope this answers the question. Have a nice day.</span>
Step-by-step explanation:
Close. You correctly set up the integrals. When integrating e²ˣ:
∫ e²ˣ dx
½ ∫ 2 e²ˣ dx
½ e²ˣ + C
So the coefficient should be ½, not 2.
[eˣ − ½ e²ˣ]₋₁⁰ + [½ e²ˣ − eˣ]₀¹
[(e⁰ − ½ e⁰) − (e⁻¹ − ½ e⁻²)] + [(½ e² − e) − (½ e⁰ − e⁰)]
1 − ½ − e⁻¹ + ½ e⁻² + ½ e² − e − ½ + 1
-e⁻¹ + ½ e⁻² + ½ e² − e + 1
Answer:
B. -4
Step-by-step explanation:
The function only has one y-value: -4. That is the entirety of its range.
The discontinuity occurs when the denominator is equal to zero, as it has "infinite" slope, and thus is not a real value or point.
x^2+x-12
x^2+4x-3x-12
x(x+4)-3(x+4)
(x-3)(x+4)
So discontinuities occur when x=-4 and x=3
