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Chikis

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  1. [math]\frac{A(x+5)+B(x+1)}{(x+1)(x+5)}=\frac{7x+19}{(x+5)(x+1)}[/math] Ok! Multiplying the left and right hand side by the denominator, we have [math]A(x+5)+B(x+1)=7x+19[/math] What happens next?
  2. [math]\frac{A(x+5)+B(x+1)}{(x+1)(x+5)}=\frac{7x+19}{(x+5)(x+1)}[/math] Could this one be right? Is the right hand side correct now?
  3. Ok, solving [math]\frac{7x+19}{(x+1)(x+5)}=\frac{A}{(x+1)}+\frac{B}{(x+5)}[/math] [math]\rightarrow[/math] [math]\frac{A(x+5)+B(x+1)}{(x+1)(x+5)}=\frac{7x+19}{1}[/math] How does this give rise to[math]7x+19=A(x+5)+B(x+1)[/math]
  4. I need help with this: I need some explanation when it comes to dealing with partial fractions. [math]\frac{7x+19}{(x+1)(x+5)}[/math] into simpler fractions.[math]\frac{7x+19}{(x+1)(x+5)}=\frac{A}{(x+1)}+\frac{B}{(x+5)}[/math] Why did[math]\frac{7x+19}{(x+1)(x+5)}[/math] become equal to [math]\frac{A}{(x+1)}+\frac{B}{(x+5)}[/math]?
  5. Alright, thank you for the asistance.
  6. Do I take it that [math]\frac{-bp}{cp}=\frac{bp}{-cp}[/math]?
  7. How do I show, prove, see or be convinced that [math]-\frac{bp}{cp-a}=\frac{bp}{a-cp}[/math] I am concerned with what is going on in the minus sign there. I do remember that the following holds: [math](-)(-)=+[/math] Or that [math](-1)(-1)=+1[/math] or just 1. I also remember that [math]\frac{(-1)}{(-1)}=1[/math] or just 1 or that [math]\frac{(-)}{(-)}=+[/math]. What can I make of these identities in being convinced or cleared of what is obscured to me.
  8. Yes, it is not factorizable. I did not notice the signs at first. I will rather have [MATH]2pq-1-p^2q^2[/MATH] which is not the original expression. But can the quadratic formular be used to solve this problem?
  9. It cancels. See it here. [MATH](1-pq)(pq-1)[/MATH] = [MATH]1(pq)+(1)(-)-(pq)(pq)-(pq)(-1)[/MATH]. This gives [MATH]pq-1-p^2q^2+pq[/MATH] Study it carefully and you will see that the pq terms cancel to give the original expression.
  10. Here is the factorization of [MATH]-1-p^2q^2-[/MATH]: [MATH](1-pq)(pq-1)[/MATH]
  11. what am saying is this, I know that the expression can be factorized so that it appears in complex number form. I also know that the expression can be factorized so that it appears in real number form. I am not intrested in making it appear in complex number form. I want to factorize it such that the product appears in real number form. What I need now from you, is a guide to make that happen. Would you mind helping me actualize my determination?
  12. The answer given is N10.65. I can't explain with details but what I will say is this:the 75k is the norminal or face value of the share. The share was bought by the shareholder below par. I.e the market value of the share is 70k. Shares are not always sold at par with their norminal value.
  13. So do you think my answer is correct without any shadow of doubt?
  14. The [MATH]i^2[/MATH] in the spoiler is an imaginery number and I don't think it is admmisible for this kind of work. I stand to be corrected.
  15. I would like to start by giving some little informations about the currency am using. N stands for Naira and k for kobo. 100 kobo = N1 Here we go: Find the gain or loss made by a shareholder who buys 90 75k shares at 70k each and sells them at 76k each. Market price for each share = CP, cost price for each share = 70k CP for 90 shares = [MATH](70\times90)k[/MATH] = 6300k = N63 Selling price for 90 shares = [MATH](76\times90)k[/MATH] = 6840k = N68.4 From want we have obtained so far, we can see that the share holder made gain. He made gain of N(68.4-63) = N5.4 My concern here is that the answer provided for the question is not saying the same thing as the answer obtained in my work. I don't know if there is anything am not doing well in this problem. Please I need help and I stand to be corrected if there is any error or poor knowledge of ideas as it has to do with this problem. Thank you.
  16. How do I factorize? [MATH]-1-p^2q^2[/MATH] I would have used the idea of difference of two squares to handle this but I found it difficult because of the negative sign. If I factor the -1, I have [MATH]-1(1+p^2q^2)[/MATH] or [MATH]-1(p^2q^2+1)[/MATH] and that makes the terms inside the bracket unfactorizable?
  17. The answer is incorrect![MATH]\frac{3.87^3+20}{3.87^3-20}[/MATH] = 2.05 I want to solve the problem using logathim and still have the same as when calculating it ordinarly.
  18. Please come to my aid. I have shown my work. I can't make any headway without help.
  19. But[MATH]\frac{3.87^3+20}{3.87^3-20}[/MATH] =[MATH]\frac{10^{(0.5877)3}+10^{1.3010}}{10^{(0.5877)3}-10^{1.3010}}[/MATH] What else can I do?
  20. [MATH]\frac{3.87^3+20}{3.87^3-20}[/MATH] I want solve the above using logarithm. I don't know how to solve it using logarithm because I can't deal with this: [MATH]\frac{10^{(0.5877)3}+10^{1.3010}}{10^{(0.5877)3}-10^{1.3010}}[/MATH] If this is given: [MATH]\frac{3.87^3\times20}{3.87^3\times20}[/MATH] It would have been easiar for me to solve if I add and subtract all the powers that has base 10. Please I need help.
  21. I think I have tried my best. Could you please come to my aid now?
  22. I think I have come up with something. From the first sentence, we know that the blend = x + y When the mixure contains 20% of x, the percentage of y in the blend was 80%. When the quantity of x is doubled, the percentage of x in the new blend becomes 40%. . (i) if the same quantity of y was used the percentage x = 40% (ii) the total quantity of the mixure = 40 + 80 = 120%, [math]\therefore[/math] x remains 40% Am also thinking that in the first sentence x/5 + 4y/5 = [math]\frac{20x}{100}+\frac{80y}{100}[/math] The ratio of x to y is 20 : 80 = 1 : 4 When x is doubled, we have 2x/5 + 4y/5 = [math]\frac{40x}{100}+\frac{80y}{100}[/math] the ratio of x to y becomes 1 : 2
  23. I think I have come up with something. From the first sentence, we know that the blend = x + y When the mixure contains 20% of x, the percentage of y in the blend was 80%. When the quantity of x is doubled, the percentage of x in the new blend becomes 40%. . (i) if the same quantity of y was used the percentage x = 40% (ii) the total quantity of the mixure = 40 + 80 = 120%, [/math]\therefore[math] x remains 40% Am also thinking that in the first sentence x/5 + 4y/5 = [math]\frac{20x}{100}+\frac{80y}{100} The ratio of x to y is 20 : 80 = 1 : 4 When x is doubled, we have 2x/5 + 4y/5 = [\math]\frac{40x}{100}+\frac{80y}{100}[math] the ratio of x to y becomes 1 : 2
  24. I have got the whole thing now? The retailer made a profit of N10 after he bought the good for N190. When he borrowed from the bank, he incured an intrest of N3. The man paid N3 out of his N10 gain. He had N7 left. His gain is N7. What if the question says what is his gain percent?
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