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Sariah
Question 19. Perform the indicated operations and simplify the result. Leave the answer in factored form. Question 19. Perform the indicated operations and simplify the result. Leave the answer in factored form. @Mathematics
Please only i need help to finish this math work,,,
denominator in the second fraction is the first denominator squared... (3x/x-1) - (x-4/x^2-2x+1) = (3x/x-1) - (x-4/(x - 1)^2) Thus we will need to multiply the first fraction by (x - 1) / (x - 1)... (3x(x - 1)/(x - 1)^2) - (x-4/(x - 1)^2) and put it all over the same denominator: [(3x(x - 1) - (x - 4)] / (x - 1)^2 Now expand the numerator and simplify... [(3x(x - 1) - (x - 4)] / (x - 1)^2 = (3x^2 - 3x - x + 4) / (x - 1)^2 = (3x^2 - 4x + 4) / (x - 1)^2 Nothing more can be done with this one...that's the answer.
Nick can you keep help me?
What are the answer to question 9,10,13,17,18 and 19...
Iam not good for math...
\[\frac{\frac{4}{x}+\frac{7}{x^{2}}}{\frac{16}{x^{2}}-\frac{49}{x}}\]first simplify numerator: \[\frac{4}{x}+\frac{7}{x^{2}}=\frac{4x+7}{x^{2}}\]next simplify the denominator: \[\frac{16}{x^{2}}-\frac{49}{x}=\frac{16-49x}{x^{2}}\]now use the following rule for fraction division:\[\frac{a}{b}/\frac{c}{d}=\frac{a}{b}*\frac{d}{c}\]and you will get: \[\frac{\frac{4}{x}+\frac{7}{x^{2}}}{\frac{16}{x^{2}}-\frac{49}{x}}=\frac{4x+7}{x^{2}}/\frac{16-49x}{x^{2}}=\frac{4x+7}{x^{2}}*\frac{x^{2}}{16-49x}\]\[=\frac{4x+7}{16-49x}\]
The D is the right answer then...
for question 9, a polynomial is defined as:\[c_0x^{0}+c_1x^{1}+c_2x^{2}+...+c_nx^{n}\] all the "c" terms are constants. it's "degree" is defined as the highest "power" of x that you have - in the case above it is a polynomial of degree "n". e.g. 2x+9x^2 is a polynomial of degree 2. you should be able to answer question 9 with this information.
question 13 is fairly easy - you just made a simple mistake when multiplying the last terms
Iam sure that Question 9 the a is not the answer right...
Please dear, give me the answers...
correct - the only term given in the polynomial in question 9 is\[-2\pi\]which is just a constant. so there are no powers of x in this.
if it is not the a, what is the right answer?
Do you want a medall? please help with the answers... I do not understand math....Iam studing othe thing....but i selected elective math and it was a wrong
I am not looking for a medal - I am here to try and you understand maths. If you are not interested in maths then I guess there is no point in me trying to explain it to you.
But if you know the answers... help me then with the answers...
Thanks dear....can i pay to you if you want... but i need answer this....
I can try and teach you how to solve these yourself. you may think its very complicated but it is fairly straight forward and it will give you greater satisfaction when you know that you can work these out yourself. I am not looking for any money or medals. Let me know if you wish to learn.
ok - good - lets start with question 9
Ok dear... where are you from?
what do you think is the "degree" of the following polynomial?\[9x-100x^{3}\] (I live in the UK)
are you thinking Sariah?
Yes...but it is very dificult...
ok - the answer here is 3 because the term with the highest power is "-100x^3". you only need to look at the powers of x - you can completely ignore the coefficients (i.e. ignore the -100 which x^3 is multiplied by)
lets try one more: what is the degree of this polynomial?\[99+3x\]
the answer is "1" because the highest power of x here is in the term "3x".\[x = x^1\] similarly, the polynomial: 123 has a degree of "0" because if has no "x"'s in it - just a constant. you should now be able to answer question 9.
If you don´t want keep helping me, don´t worry! But thanks anyway...Really for me this assignment is dificult... you are a very good person....´couse i don,t have idea...
I think the question 9 is D
ok Sariah I need to go and eat something now - good luck with the future and be more positive in your own abilities!
Ok dear... Have a nice day...
sure - it's openstudy@anjum.otherinbox.com by the way - just so that you don't get the wrong impression - I am a VERY happily married man :-)
Ok Taht is good....Good luck! If i have question in the fucture, can i send you the E-mail to you? or what do you think?
no problem - got to go now - should be back here in about 2 hours....
I see you are very good man... i wish every one be like you... Bye Bye!
Not today maybe but in the end of november... is it ok... or one day of this.. bye and keep being so you are