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I really don't understand hypothesis testing. How do I recognize which one is Ho and Ha in a problem? Is there a particular trick to facilitate my lfie?

How do we find the parametric equations for the line of intersection of the given planes 7x-2y+3z = -2 and -3x+y+2z+5=0 :approve:

Thanks, it is appriciated.

How can i proove if this line is perpendicular to this one? Line x=-2-4y, y=3-2t, z=1+2t Plane 2x+y-z=5 I don't care that much about the answer, I want the procedure.

(e+f)^2 = e^2+2ef+f^2 Why is this not neccesary true for squares matrices for all size?

Thanks bud, i'll look at it later on. If i have any problem, i'll post.

At first, i putted the first number in the frist row as 1. so i get [1 1/3 1/3 1/3 0] than i eliminate the 5 to make it a 0. But its after that i got ****ed! mayday!

Linear stuff: I never get the awnser...! [3 1 1 1 0] [5 -1 1 -1 0] Where the variables are X1 X2 X3 X4. I always get somethignt that is far away from the answer. The answer should be x1= -s x2=-t-s x3=4s x4 = t Help please!

Ok, I have to integrate this [int a=0 b=3] 1 / sqrt[abs(x-2)] dx what should i do ? do the improper integral of 1/-(x-2) and 1/(x-2) ?

In my paper it says use disk method. :cry:

I don't get it :confused:

I don't know if I did good, cause I'm uncapable of integrating it. :cry:

[int a=0 b=1] pie(arcsiny)^2 dy disk method.

How do I find the indefinite integral of 1/[(e^x)+(e^-x)] ? Do I have to multiply by [(e^x)-(e^-x)]/[(e^x)-(e^-x)] to eliminate the denominator? !

Can anybody help me find the volume y=sinx, bounded by the y axis, and the line y=1. The axis of revolution is the line y=1. I tried so many times and I can't find the correct answer, in the sheet it says (pie^2)/2 - 2pie

but in my problem, it must have respect to y... How about the reciprocal of x=2, would it be y=2 ? Just swapping the variable.

How to we do the inverse of y=(x-1)^2 ? Would it be x = sqrt(y) +1 ?

What would the inner and outter radius of y=arccos(x)

Answer sheet = 64pie/15 But I think the paper is wrong, cause it keeps giving me 64pi/12 (Unreduced)

Thx man, I gotted the answer but I was too tired to realize it :rofl:

I did the reciprocals. so its x= sqrt(x)+1 and x=y/2 +1 I did everything and the result is 16pie/12, and in the answer sheet it says 64pie/12

R is bounded by the curves y=(x-1)^2 and y=2(x-1). Axis of revolution: y-axis. How am i supposed to do this. I know how to do the washer method, I know how to apply it when it is revolved arround the x-axis, but I don't know how to do it when it is the y axis. Can anybody explain to me the...

I don't know man, maybe the answer sheet is wrong. It says 9pie/4,as I mentionned before. I tried, I had pie/2, and test it on graphmatica the program, and it gave something near it.

I did the thing, and it still gives me 4 :cry:

Here's a integral where I have to use trigonometric substitution but I can't get the right answer. [int a=0 b=3] 1/(sqrt[9-x^2]) dx I did the limit as t approches 3 from the left. Then i did my trigonometric substitution, and it gives me arcsin(x/3). Then i computed what i had...

The limits would be 1+ and 1- right ?

Let's say something like [int a=0 b=2] 1/(x-1)^(1/3) dx

But the limit for that problem have to approch 2 from the right for the x-2, and from the left for -(x-2). Right ?

Nice, i check it out on gragphmatica, and the graph of both seems to match the original function. Thanks dude, if i have a question i<ll comme back here :tongue:

I tried and it keeps giving me 0. 2[sqrt(abs(x-2))] from 1 to t and 2[sqrt(x-2)] from t to 3 (2[sqrt(abs(1-2))])-(2[sqrt(abs(t-2))]) and (2[sqrt(abs(t-2))])-(2[sqrt(abs(3-2))]) (lim t->2+ (2[sqrt(abs(t-2))])-(2[sqrt(abs(1-2))]) ) + (lim t->2-...

I've posted on Homeworks one of the number I did not understand. However, I would like to know the steps to calculate an improper integral of type 2. The type 2 is the one from constant a to constat b, not the one with inifnite. Please tell me the steps the accomplish it. :smile: I know...

I don't understand my Math Homework, here's the number that I don't get. [int a=1 b=3] 1/(sqrt[abs(x-2)]) dx sqrt = square root abs = absolute value Integral from 1 to 3 Can anyone explain this to me clearly, I'll really appreciate :smile: Thanks