These do start with 41 and 41^2 instead of 1 and 1^2, but the formula they have (the ##\frac{x(x+1)}{2}##) is for the sequence that starts with 1.
So that means that you can rewrite the formula given as a sigma notation from 1 to n and then simplify from there.
Homework Statement
If r is a rational function, use Exercise 57 to show that ##\mathop {\lim }\limits_{x \to a} \space r(x) = r(a)## for every number a in the domain of r.
Exercise 57 in this book is: if p is a polynomial, show that ##\mathop {\lim }\limits_{x \to a} \space p(x) = p(a)##...
That is true if it is just the quadratic equation, but if you are dividing two numbers what happens if they are each positive? Each negative? One positive and one negative?
You need to use the answers to those questions to determine the sign of the division of two functions.
I don't know if you want a mathematical method for doing a problem like this, but I would approach it by saying that if a person speaks only English, than they don't speak Spanish or Swahili.
So there are 100 people, and at least 60 of them don't fit in the set of people who only speak English...
It does effect the acceleration of the mass system because as the mass drops, energy is transferred into the pulley as well, instead of remaining in the system.
Ah okay :)
You said that you get the correct answer when you use the formula with a ##2g## in the denominator instead of just ##g##, and when you use the ##g## formula you get it wrong.
But when you use the formula with just a ##g##, did you recalculate your ##V_o##?
I'm confused, if you are able to hit a ball the same distance with the same stroke on two different planets, wouldn't that mean that gravity was the same on each?
So you want to take the square root of each side.
If ##y^2 = 1 - x^2## then ##\sqrt{y^2} = \sqrt{1 - x^2}##, which is only the same as ##y = 1 - x## if ##\sqrt{a + b} = \sqrt{a} + \sqrt{b}##.
Well if you call heading east to be positive, vi will be positive, Δp will be negative, and your result will be 4.2
If you call going west to be positive, vi will be negative, Δp will be positive, and your result will be -4.2 (and negative refers to going east).
So the answer will have the...
Remember with absolute value that it leads to two separate equations; |f(x)| = g(x) leads to f(x) = g(x) or f(x) = -g(x),
This also applies to inequalities.
|x| < a is the same as x < a or x > -a, which is condensed into -a < x < a.
Similarly, |x| > a is the same as x > a or x < -a.
With...
Your impulse and your velocity are in opposite direction, so you have to make one or the other negative in order to reflect this.
I haven't checked to see if this yields the correct answer based on the rest of your work, but it should.
However, I think that it would be simpler if you use...
If you're using latex, you type \cdot.
Otherwise you can't, but in order to avoid confusion you can write (√3)y to illustrate the difference.
But before we can help you figure out the problem, you need to show the work you already did.
I'm not sure if the part I bolded is true, I've never heard it before.
Where were you told this?
The possible domain for fg(x) is the range of g(x), which is true. But that possible domain can still be further restricted based on the domain of f(x).
For this problem I would just look at the final answer for gf(x) because ##\sqrt{x-3}## is a simple solution.
But are you more interested in a general way to doing these problems?
Show your attempt, maybe we can help you figure out what went wrong.
Based on the answer he gave, I would think that it has to be the first equation that you listed.