Answer:
sin−1(StartFraction 8.9 Over 10.9 EndFraction) = x
Step-by-step explanation:
From the given triangle JKL;
Hypotenuse KJ = 10.9
Length LJ is the opposite = 8.9cm
The angle LKJ is the angle opposite to side KJ = x
Using the SOH CAH TOA Identity;
sin theta = opp/hyp
sin LKJ = LJ/KJ
Sinx = 8.9/10.9
x = arcsin(8.9/10.9)
sin−1(StartFraction 8.9 Over 10.9 EndFraction) = x
Answer:
Step-by-step explanation:
Assuming this integral:

We can do this as the first step:

Now we can solve the integral and we got:

![\int_{-\infty}^0 5 e^{60x} dx = \frac{e^{60x}}{12}\Big|_{-\infty}^0 = \frac{1}{12} [e^{60*0} -e^{-\infty}]](https://tex.z-dn.net/?f=%20%5Cint_%7B-%5Cinfty%7D%5E0%205%20e%5E%7B60x%7D%20dx%20%3D%20%5Cfrac%7Be%5E%7B60x%7D%7D%7B12%7D%5CBig%7C_%7B-%5Cinfty%7D%5E0%20%3D%20%5Cfrac%7B1%7D%7B12%7D%20%5Be%5E%7B60%2A0%7D%20-e%5E%7B-%5Cinfty%7D%5D)
So then we see that the integral on this case converges amd the values is 1/12 on this case.
It doesn’t make any difference whether or not the data is grouped, the mean is the same. So b. false.
To solve this
First I would change the mixed number on the right side of the equation to an improper fraction
7/2 x=-5
Then I would multiply both sides by 2/7
x=-10/7
16 because the neg next to the subtraction sign becomes positive then you’re just adding 14+2.