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#526 Re: Help Me ! » Simplify the following: » 2013-07-10 08:07:36
Err sir I have understood you well! with (1) and (2). :-)
#527 Re: Help Me ! » Simplify the following: » 2013-07-10 08:01:48
= 4m^2 + 6mk + 6mk + 9k^2 - m^2 - 2km - 2km + 4k^2
There is a sign error here. The -2km terms must be subtracted so they become +
Bob
Please I am not grasping what you mean could you demonstrate? Or you mean:
6km + 6km - 2km - 2km = 12km - 4km = 8km
#528 Re: Help Me ! » Simplify the following: » 2013-07-10 07:51:45
I see it the following:
Difference of two square: (p^2 - q^2) > I suppose it must be something like that in other to be called difference of two square, but I think something like (2m + 3k)^2 should be opened before manipulation, as I have done above.
Something like (q - p)^2 I think is not a difference of two squares. The book's final answer to the first question is (3m+ k)(m +5k). That really confuses me.
#529 Re: Help Me ! » Simplify the following: » 2013-07-10 07:39:31
(2m + 3k)^2 - (m - 2k)^2
= (2m + 3k)(2m + 3k) - (m - 2k)(m - 2k)
= 4m^2 + 6km + 6km + 9k^2 - m^2 - 2km - 2km + 4k^2
=3m^2 + 8km + 13k^2.
That is it! How correct I am?
Thanks.
#530 Re: Help Me ! » Simplify the following: » 2013-07-10 07:23:23
Yes, with the difference of two square I have met that before.
But with respect to the first problem, my final answer was inconsistent with that of the book. But I I would try here.
#531 Re: Help Me ! » Simplify the following: » 2013-07-10 07:09:21
Please I need an assistance here:
Factorise the following expressions completely,
(1) (2m + 2k)^2 - (m - 2k)^2.
(2) ( x + 3)^2 - (x-7)^2.
(3) 9p^2 - (q^2 - 6qr + 9r^2).
(4) 4a^2 - 12ab - c^2 + 9b^2.
In each case the answer has been provided, but the procedures wasn't shown.
Thanks in advance.
#532 Re: Help Me ! » Simplify the following: » 2013-07-10 06:54:40
No sir, I cannot even locate where they are!
#533 Re: Help Me ! » Simplify the following: » 2013-07-09 06:33:00
Thanks, anonimnystefy.
But one thing left, I tried to use the SYMBOLS, I mean the mathematical symbols as you have used above nicely to solve the problem, but I couldn't locate where they were. Could you please assist me to find them and use them each time I want to post? The moderator, provided the link in my previous post, but I couldn't use them in this particular post.
Thanks. ;-)
#534 Help Me ! » Simplify the following: » 2013-07-08 08:57:54
- EbenezerSon
- Replies: 713
Simplify the following.
1/x-3 - 3(x-1)/ x^2-9
I have worked out this, but my final answer was different from what is at the back of the book.
The book has given the final answer as:
-2/x-3
The book did not provide the procedures, just the answer, please help.
Thanks in advance.
#535 Re: Help Me ! » convert sum od exponents into product of exponents » 2013-07-08 08:44:05
Is it an indicial equation?
#536 Re: Help Me ! » Apply logarithms to both sides of an equation » 2013-07-08 03:30:07
Yes I did it as an example, It not the real values.
Thanks.
#537 Re: Help Me ! » Apply logarithms to both sides of an equation » 2013-07-08 03:18:25
See the following:
Solve the following logarithm:
Log6 + Log10. Given that, Log2=3.14 Log3=616
= Log(2*3)+Log10 = Log2+log3 +log10= (3.14+616+1) = 620.14.
Note that, Log10=1
This is what I mean.
I hope you cotton.
#538 Re: Help Me ! » Apply logarithms to both sides of an equation » 2013-07-08 03:05:03
whe will a log of a number return a whole number?
Sometimes the when a log question is given, one could arrive on figures which you must substitute them into it. Actually the question setter, will set it in such way that you will arrive on those whole numbers. An finally you will substitute.
#539 Re: Help Me ! » Apply logarithms to both sides of an equation » 2013-07-08 02:54:05
and if I had something like log(3^3+3^4)=log(3^7), how could I simplify log(3^3+3^4). How could you solve this, because I tried adding what's inside the parenthesis and then applying logarithm and it didn't work. How can I solve this equation?
Check very well from the source I suppose there should be some kind of: log2=3.142 or Log3=616 or something, so that you can substitute into the equation, because to me this log could be solved to some extent but good final solution is not possible. If something of that kind is not provided.
#540 Re: Help Me ! » Algebraic problem. » 2013-07-08 02:33:55
Hi;
That is an extremely weird answer they give.
Sir, forgive me I should have provided how the question should be solved. the question says:
(The factorize the expression completely). I think it should be the answer if we are to factorize it, I have now comprehended it. And I can explain.
What do you say.
#541 Re: Help Me ! » Algebraic problem. » 2013-07-08 02:20:46
(h^2 - h^2) -p(h+h)
= (h - h)(h + h) -p(h+h)
= (h + h)(h - h - p) . The final solution provided by the book.
#542 Re: Help Me ! » Algebraic problem. » 2013-07-08 02:12:39
Regarding the problem above I think the member omitted the 'negative sign'.
The following is how I see it;
0-p(h+h) = (-ph - ph)
= -2ph , What do you say with my solution? The 'negative sign' was affecting the P.
#543 Re: Help Me ! » Algebraic problem. » 2013-07-05 10:44:02
But sir, the book in which I am learning this from has given me a different answer altogether.
It is saying, though an algebra but this type of question is called "difference of two squares".
I would provide his procedure here, which I could not quite comprehend some aspects of his procedure
.
I really appreciate your enormous assistance.
Thanks.
#544 Re: Help Me ! » Algebraic problem. » 2013-07-05 10:38:38
0-p(h + h)
= -hp - hp.
#545 Introductions » My greetings » 2013-07-05 04:21:30
- EbenezerSon
- Replies: 6
Speaking candidly, I am not quite good at introduction, this is the little I would say;
Yeb, I really enjoy maths a great deal but through understanding, I was surfing the net as I chanced on this site.
I really smiled having found this forum, cos I would learn more stuff here, Ironically my friends really detest math while I love it. They always come to me for assistance at areas they feel incapable to solve.
But one thing is that I like asking questions especially when I am confused, so please I know you're kind persons to assist me with my mathematical problems.
I would tend to say that, as I am a member here, I am delighted and I know that my problems would be no more. Lest I forget, I was motivated the more, by the caption of the kind moderator. thanks. :-)
God bless all.
#546 Re: Help Me ! » Algebraic problem. » 2013-07-05 03:57:54
Thank you sir, God bless, but please one more problem from algebra.
1. (h^2 - h^2) -p(h + h).
2. X^2 + X^2 + X^2 - Y^2
Please, where can I use the mathematical items, like the (square), (plus symbol) and the others in this forum, as you have used them nicely above.
Thanks.
#547 Help Me ! » Algebraic problem. » 2013-07-04 11:03:57
- EbenezerSon
- Replies: 13
Hello sir, please assist me to solve these algebra.
x+x+x2-x2. please the two x are squared, I cannot locate them(the squares) in this forum in order to apply them conveniently on them.