Answer:
Correct integral, third graph
Step-by-step explanation:
Assuming that your answer was 'tan³(θ)/3 + C,' you have the right integral. We would have to solve for the integral using u-substitution. Let's start.
Given : ∫ tan²(θ)sec²(θ)dθ
Applying u-substitution : u = tan(θ),
=> ∫ u²du
Apply the power rule ' ∫ xᵃdx = x^(a+1)/a+1 ' : u^(2+1)/ 2+1
Substitute back u = tan(θ) : tan^2+1(θ)/2+1
Simplify : 1/3tan³(θ)
Hence the integral ' ∫ tan²(θ)sec²(θ)dθ ' = ' 1/3tan³(θ). ' Your solution was rewritten in a different format, but it was the same answer. Now let's move on to the graphing portion. The attachment represents F(θ). f(θ) is an upward facing parabola, so your graph will be the third one.
Answer:
x=-1
Step-by-step explanation:
x+2-1=0
-1+2=1
1-1=0
y=0
Answer:
0.833 feets
Step-by-step explanation:
Given that:
Volume of box = 1600
Heigh of box (H) = 8 inches
Let:
Width of box (W) = x
Length of box (L) = 2x
Volume of box = (Height * width * length)
1600 = (8 * x * 2x)
1600 = 16x²
x² = 1600/16
x² = 100
x = sqrt(100)
x = 10 inches
Hence, width of box in feet:
1 inch = 0.0833 feets
10 inches = (10 * 0.0833) feets
= 0.833 feets
<h2>
Answer with explanation:</h2>
We know that a removable discontinuity occurs when:
The left and the right hand limit of the function exist at a point and are equal but is unequal to the function's value at that point.
Also it is a point on the graph such that it is undefined at that point.
The graph that has a removable discontinuity is attached to the answer.
Since, at x=0 the left hand and the right hand limit of the function exist but the function is not defined at x=0 , since in the graph there is a open circle at x=0 that means that the point is removed from the range.
Answer:
15
Step-by-step explanation:
195/13=15
