Answer:
You should use photomath, it would be more easier for u. Oh, and by the watñy Y=12
The best and most correct answer among the choices provided by the question is the third choice .
In constructing a parallel line, "<span>Without changing the width of the compass, place the compass at S or P and draw an arc similar to the one drawn."</span>
I hope my answer has come to your help. God bless and have a nice day ahead!
Answer:
-8+0
Step-by-step explanation:
Answer:
![-1.9338](https://tex.z-dn.net/?f=-1.9338)
Step-by-step explanation:
Answer:
INF for first while D for second
Step-by-step explanation:
Ok I think I read that integral with lower limit 1 and upper limit infinity
where the integrand is ln(x)*x^2
integrate(ln(x)*x^2)
=x^3/3 *ln(x)- integrate(x^3/3 *1/x)
Let's simplify
=x^3/3 *ln(x)-integrate(x^2/3)
=x^3/3*ln(x)-1/3*x^3/3
=x^3/3* ln(x)-x^3/9+C
Now apply the limits of integration where z goes to infinity
[z^3/3*ln(z)-z^3/9]-[1^3/3*ln(1)-1^3/9]
[z^3/3*ln(z)-z^3/9]- (1/9)
focuse on the part involving z... for now
z^3/9[ 3ln(z)-1]
Both parts are getting positive large for positive large values of z
So the integral diverges to infinity (INF)
By the integral test... the sum also diverges (D)