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#176 Re: This is Cool » Something ineteresting » 2018-10-09 23:34:30

Hi Βεν,

Welcome to the forum! Thanks for your contribution. That looks like a nice list. Some comments:

a^n=a multiplied by itself n times

I know what you mean, but this can be misleading:
is multiplied by itself
times. For example,
, where
gets multiplied by itself one time.

If n is even then (-a^n)=a^n

The brackets should go around the
here, i.e.
for even
.

a^n÷a^m=a^(m-n) (Makes sense right?)

On one side of the equation, the
and
are the wrong way round. It should read:

If n>0 then (a^m)na=a^(m+n)

What did you mean here?

#177 Re: Euler Avenue » Sine and cosine » 2018-10-03 01:13:18

Hi zahlenspieler,

I have fixed your LaTeX.

#178 Re: Help Me ! » Algebra: Polynomial Question » 2018-10-01 01:12:02

Hi math9maniac,

Can I just check that the equation you wrote down is correct?

#180 Re: Help Me ! » Exponential expression » 2018-09-28 22:32:53

OK, suppose we had
. Then:

In this case, we have
. We want to calculate
, which means we need to multiply the index (
) by
. Can you see how to do this?

#181 Re: Help Me ! » Exponential expression » 2018-09-28 21:14:39

Hi vipin_sharma,

Welcome to the forum. Here's a hint:

#182 Re: Help Me ! » answer for this series: 4, 4, 341, 6, 4, 4, 6, 6, 4, 4, 6, 10, 4, 4, 1 » 2018-09-23 05:52:10

Hi niravashah,

Thanks for posting this. Have you considered creating an account with us?

For reference, the sequence is:

4, 4, 341, 6, 4, 4, 6, 6, 4, 4, 6, 10, 4, 4, 14, 6, 4, 4, 6, 6, 4, 4, 6, 22, 4, 4, 9, 6,

and we're asked to find the next two terms. Perhaps this will give you a hint:

4, 4, 341, 6, 4, 4, 6, 6, 4, 4, 6, 10, 4, 4, 14, 6, 4, 4, 6, 6, 4, 4, 6, 22, 4, 4, 9, 6,

#184 Re: Help Me ! » Conditional Statements and Venn Diagrams » 2018-07-28 04:59:30

Draw a Venn diagram with 3 overlapping circles, labelling the four bits that overlap as A, B, C, D.

The information given in the question then allows you to form some equations for A, B, C and D. Can you see how?

#185 Re: Help Me ! » Find all real solutions: Proof help » 2018-07-27 21:23:51

!nval!d_us3rnam3 wrote:

Ok, but I wasn't looking for a proof with the answers in the beginning. I've gotten pretty far, but I need to find all the answers to


EDIT: With a rigorous demonstration that this is the answer.

That depends on what you consider to be a 'rigorous demonstration'. Clearly
is a factor, since both the LHS and RHS vanish. Dividing your equation by
(taking
) gives you

Since
vanishes whenever
, then the numerator of the term on the RHS also vanishes for these values of
. That's three solutions, and you can 'rigorously demonstrate' that there aren't any more solutions over
via the method in post #4.

#187 Re: Introductions » Hi there. » 2018-07-03 03:03:20

Hi Stuti55,

Welcome to the forum.

#188 Re: Help Me ! » Division » 2018-07-02 20:30:53

Hi Zeeshan,

I've moved your thread to the Help Me ! forum.

Have you checked out Maths Is Fun's page on long division?

#189 Re: Help Me ! » Precalculus Help » 2018-06-19 09:28:17

Hi Anduin,

Welcome to the forum. What features does the graph y = sin(x) have? How do they compare to the graphs y = 2sin(x) and y = 3sin(x)?

#190 Re: Help Me ! » algebra » 2018-05-27 22:57:19

Hello,

I noticed that you've posted several times here as a guest. Why not register an account with us?

What have you tried? What sorts of restrictions do you think you can place on x and y?

#191 Re: Help Me ! » Algebra » 2018-05-07 08:27:45

Grantingriver wrote:

Hi Zetafunc, if there are duplicate roots of a polymolial you should counts them as distinct roots.

Why? Precisely the opposite is true: if a root is repeated, then we cannot call those repeated roots 'distinct', and whether or not we count them in the same way we would distinct roots depends on the context.

Grantingriver wrote:

For example, if you say that a give polynomial has “exactly” the roots -3, 4 and 8 that means there is no duplicate roots.

I am not convinced that the word 'exactly' necessarily implies distinctness -- I would have thought that the multiplicities were worth consideration, otherwise the problem appears to be a little simplistic.

#192 Re: Help Me ! » Algebra » 2018-05-05 20:41:44

Grantingriver wrote:

The answer for part 1 is, by all means, yes. This is so because the condition that f(x) has the zeros -3,4 and 8 entails that:
f(x)=(x+3)×(x-4)×(x-8)
And since the zeros of g(x) are -5,-3, 2, 4 and 8 we also have:
g(x)=(x+5)×(x+3)×(x-2)×(x-4)×(x-8)
and hence:
g(x)/f(x)=[(x+5)×(x+3)×(x-2)×(x-4)×(x-8)]/[(x+3)×(x-4)×(x-8)]=(x+5)×(x-2)
and since the result of the division (x+5)×(x-2) is a polynomial without a remainer, therefore g(x) is divisible by f(x).  The answer of part 2 is very clear from the answer of part 1. For a polynomial g(x) to be divisible by a polynomial f(x) at least all the roots (zeros) of f(x) should be also roots (zeros) of g(x). And the final statement completes the answer.

This is not quite true: the multiplicities of the zeroes are important, e.g.
  and 
.

#193 Re: Help Me ! » Help Me ! » 2018-04-30 06:42:02

Hello bishop.bell,

Welcome to the forum. This problem has been asked at least twice in the past -- you can find some discussions about it here and here.

#194 Re: Help Me ! » a question of a dice in probabilty » 2018-03-30 21:10:17

The denominator will not be 36. Remember: you want to restrict the number of possible outcomes to include only those that have a '5' in them.

#195 Re: Help Me ! » [ASK] Integral » 2018-03-18 05:42:18

The question as it stands has no answer (the area would be infinite). There is probably a typo in the question.

#196 Re: Help Me ! » [ASK] Integral » 2018-03-18 04:54:32

The integral should evaluate to
. Your arithmetic slips take place in these three lines:



In particular, the second pair of brackets in the second line and the first pair of brackets in the third line are not correct.

However, the question appears to be badly worded. The area 'bounded' by the curve, the axes and the line x = 2 is not bounded at all -- it is infinite.

#198 Re: Help Me ! » [ASK] Professional Goalkeeper » 2018-03-17 21:41:25

Yes -- do you know how to calculate this?

#199 Re: Help Me ! » [ASK] Professional Goalkeeper » 2018-03-17 19:34:15

One way is to use the binomial distribution.

Suppose that
represents the number of times that the goalkeeper saves a penalty. Then,
has a binomial distribution:

You want to calculate
. Do you see why?

#200 Re: Help Me ! » Probability Statistics: Please help! » 2018-03-14 00:18:34

Hi MellyBigD,

Welcome to the forum. You have posted 4 threads asking for help, but you have not shown any attempts at the problems or given any pointers about where you are stuck. We will endeavour to help you as best you can, and we happily volunteer our time to do so, but we will not do your homework for you. Can you please show some attempts at these problems, or be specific about what exactly you are stuck on?

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