Answer:
∫₂⁵ ln(x) dx
Step-by-step explanation:
lim(n→∞) ∑ᵢ₌₁ⁿ (3/n) ln((2n + 3i) / n)
lim(n→∞) ∑ᵢ₌₁ⁿ (3/n) ln(2 + (3/n) i)
The width of the interval is b−a = 3, and there are n rectangles. So the width of each rectangle is 3/n, and the height of each rectangle is ln(2 + (3/n) i).
The ith term is 2 + (3/n) i, so a = x₀ = 2. Therefore, b = 2+3 = 5.
So the region is the area under f(x) = ln(x) between x=2 and x=5.
∫₂⁵ ln(x) dx
Answer:
The answer is C.
Step-by-step explanation:
Given formula h(t)=−16t2+v0t+h0 , where v0 is the initial velocity and h0 is the initial height.
In this case, the initial postion is a platform 30ft above ground so h0=+30
The initial velocity is 38 ft/s straight up into the air so v0=+38
h(t)=-16t2+38t+30
When the object hits the ground, h=0.
h=-16t2+38t+30=0
Simplifying 8t2-19t-15=0
(8t+5)(t-3)=0
t=-5/8 or 3
As time cannot be -ve, t=3s. The answer is C.
Answer:
The slope of the line is m = 3
Step-by-step explanation:
You can easily verify that if x decreases by 2 (e. g., from -4 to -2), y increases by 6. Thus, the slope is m = rise/run = 6/2 = 3
2004 was a leap year so 29/366=7.9%
