F = t ⇨ df = dt
dg = sec² 2t dt ⇨ g = (1/2) tan 2t
⇔
integral of t sec² 2t dt = (1/2) t tan 2t - (1/2) integral of tan 2t dt
u = 2t ⇨ du = 2 dt
As integral of tan u = - ln (cos (u)), you get :
integral of t sec² 2t dt = (1/4) ln (cos (u)) + (1/2) t tan 2t + constant
integral of t sec² 2t dt = (1/2) t tan 2t + (1/4) ln (cos (2t)) + constant
integral of t sec² 2t dt = (1/4) (2t tan 2t + ln (cos (2t))) + constant ⇦ answer
Answer:
13, 8, 3
Step-by-step explanation:
Subtract 5 from the following terms.
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Probaility in general is defined as the ratio of positive outcomes over the total number of outcomes.
In the first example, the total outcomes are 16; let us count the positive ones. There are 8 even numbers from 1-16. The prime numbers are 2,3,5,7,11,13. Out of those, only 5 are odd. Hence, in total there are 13 positive outcomes. Thus, the probability is 13/16=81.25%
Let's restrict the problem to the students that studied for the exam; the proportion is 0.57 of the total students. 0.52 of the total students studied and saw an increase in their exam. Hence, the probability that a student who studied saw an increse is 0.52/0.57 (here a positive outcome is the proportion that saw an increase and the total outcomes are all the students that studied). 0.52/0.57=91.22%
You distribute the -4 with the -8x and -8 which results in 32x+32.
Answer:
Step-by-step explanation:
r=(2,-3,0) + t (1,0,-1)
r=(2+t, -3+0t,0+(-t))
x(t)=2+t
x(t)=t+2
y(t)=0t-3
y(t)=-3
z(t)=0+(-t)=-t
z(t)=0-t
z(t)=-t
