Answer:
3/4
Step-by-step explanation:
36/48
Divide the top and bottom by 12
36/12 = 3
48/12 =4
36/48 = 3/4
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.
Square ABCD and square EFGH will reflect onto themselves across 8 lines of reflection
Answer:
this ven diagram is 4 u lol
Step-by-step explanation:
First find a common denominator and combine the fractions in the numerator:

Now simplify and cancel out all the terms that you can:

Since the remaining expression is continuous as a function of

, you can directly substitute to end up with