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Thanks, both of you, I was able to get the awnser from your help.

Hello, I was working on a math assignment and I got stuck on these two questions: 7. Solve: Sqrt(2x+25) - 2 Sqrt(x+4) = 1 I know the answer is 0 but I don't know how to remove the Square roots without ending up that x somehow equals 20. 8. Solve: { x +2y =16 { {...

Hey, I've been doing this pre-calculus assignment and I ran into this Question: "Find the equation of the set of all points (x,y) such that the absolute value of the difference between their distance to (-3,0) and to (3,0) is 4." What do they mean? I'm certain I can do it but I have no idea...

Hi, I'm having a huge amount of trouble with this question: "Write the equation of an ellipse centered at the origin that has a horizontal major axis e = 1/2 and 2c=1" First off, what is "2c=1" I have never seen this before, is it telling me that c = .5? that makes no sense though, because e =...

I told you what methods I was taught, none of them work. I actually know the answers already because I used a graphing calculator, but I have to show your work full points, but whatever then, it's only one question

Hi, I was given the following question and I've been working on it for hours and I have no idea how to solve it due to it having no Rational roots. Find all possible roots to the nearest tenth for: 6x3+6x2+15x-2=0 Can I get some advice on where to even start? I tried Synthetic Division, long...

So I didn't notice that 2a-2 =2(a-1) would've saved me some work but I came to the same conclusion. there is no solution. thank you.

Hi, I was given the following question: Solve the equation. Check your Solution. \displaystyle{\dfrac{1}{2a-2}=\dfrac{a}{a^2-1}+\dfrac{2}{a+1}} So I get the LCD: (2a-2)(a+1)(a-1) Start canceling...

Really, that's it, I just write it out? I don't need to factor and cancel stuff and all that

Hi, I've been doing Pre-Calculus by Correspondence for a little while now, and after a while everything was going fine, but now I'm having trouble with the following composite Function question: F(x) = (3x)/(x^2 + 3x + 2), g(x) =(5x)/(x+1) simplified expression for (f•g)(x), I get about as far...

OHHHHHHHHHHHH (x2+1)2 is the same as (x2+1)(x2+1) Which would be: (x4+2x2+1) I completely forgot. I do tend forget basic math when doing this kind of stuff early on. Thank you. I never thought that as long as the top part is 0, it doesn't matter what the bottom part of the fraction is...

Thanks, that actually made me notice an error, since I screwed up the power rule (somehow it slipped my head but anyway): u(x) = x2-1 u'(x) =2x v(x) = x2+1 v'(x) = 2x therefore: f′(x)=[(2x)(x2+1)−(2x)(x2-1)]/[(x2+1)]2 f′(x)= (4x)/(x4+1) My problem wasen't that is it was: 0=(4x)/(x4+1)...

Hi, I'm sorry but I need help with calculus again. I've been doing pretty good ever since I started doing graphing, but this question is has got me stumped as what to do next: find the intervals of increase. use the first derivative test to find the local maxima and minima. sketch a rough...

So I was right then? huh, okay, thanks.

So I got this sequence in my homework and I just don't understand what I'm supposed to do since it's unlike any sequence I've ever seen. the question is as follows: a1=4, a2=3, an=3an-1+2an-2, n>2, n∈N Find the first 5 terms. at first I thought an=3an-1+2an-2 would work so that for say a3...

Exactly, that's what I meant. 3, 3+1 is 4, etc. But I have a hard time expressing and simplifying it.

Where did Come from in your answer?

but wouldn't -2 make all my answers negative?

So thanks to you I figured out this: T3(1+r)=36/t3r2(1+r)=144 = r2=4 r=2 then I did 36/3=t3 t3=12 so.. 12/2 = t2 =6 6/2= t1= 3 thus t1=3 t2=6 t3=12 t4=24 t5=48 t5=96 But the question says, (as stated before) sequences So I'm not sure what to do now?

Yeah... I know it's sad, but I don't understand it, I'm doing math by correspondence and I don't have anyone to help me with it, I've watched up like 10 videos and at least read four guides but I still don't get it. I've been in this calculus for 2 weeks now and I figured I skip it and coast...