Find A & B

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How about one more tonight?
Is this an important skill to have for Calculus ll?

20211119_212538.jpg
 
yes it is, everything in math is important

Algebra I skills are crucial to later math courses, and you must master skills like solving systems of equations, graphing, slope, and simplification of radicals. Even in Calculus, most problems consist of one difficult step, followed by ten steps of Algebra.
 
yes it is, everything in math is important

Algebra I skills are crucial to later math courses, and you must master skills like solving systems of equations, graphing, slope, and simplification of radicals. Even in Calculus, most problems consist of one difficult step, followed by ten steps of Algebra.

I can do this. I will try first.
 
MSP36031ii1369h1dbibdcf00004g696g82cb9975a6


restricted values of x are: x=3, x=-3

MSP558111agdhe67ec8g4ie0000560hf4ba0a4hie11


denominators same, so we have

MSP499119833if2ff202ie90000381da586hdh3ba19


substitute x=3

MSP6002131664i0h924927900000f9g83iechg13ia1


MSP14141hieg278hc471e000065b88c00g93eg8e9


6B=42
B=7

substitute x=-3

MSP4092198341498d7f987b00004c696b3782g790ea

MSP69741gigfcb41f95e0h900003f9hg9cf8hfg6g9d

12=-6A
A=-2


MSP10371gigg08b6i6651f40000627fbgg30i56d1fe
 
MSP36031ii1369h1dbibdcf00004g696g82cb9975a6


restricted values of x are: x=3, x=-3

MSP558111agdhe67ec8g4ie0000560hf4ba0a4hie11


denominators same, so we have

MSP499119833if2ff202ie90000381da586hdh3ba19


substitute x=3

MSP6002131664i0h924927900000f9g83iechg13ia1


MSP14141hieg278hc471e000065b88c00g93eg8e9


6B=42
B=7

substitute x=-3

MSP4092198341498d7f987b00004c696b3782g790ea

MSP69741gigfcb41f95e0h900003f9hg9cf8hfg6g9d

12=-6A
A=-2


MSP10371gigg08b6i6651f40000627fbgg30i56d1fe

I thank you very much. However, I did promise to try solving for A and B myself. Please, keep in mind that I work 40 overnight hours. Sleeping during the day is not easy at all. I look like a zombie most of the time.

The weekend hours fly by. Life often gets in the way. I am not lazy. I am committed to improving certain math skills, if not all. I just need time but time runs away from me, from all of us.
 
Another way to do this is to actually add the fractions on the right. $\frac{A}{x+ 3}+ \frac{B}{x- 3}= \frac{A(x- 3)}{(x-3)(x+ 3)}+ \frac{B(x+ 3)}{(x- 3)(x+ 3)}= \frac{Ax- 3A+ Bx+ 3B}{(x- 3)(x+ 3)}= \frac{(A+ B)x- 3A+ 3B}{(x- 3)(x+ 3)}$.
As far as "Calculus II" is concerned, t
 
Another way to do this is to actually add the fractions on the right. $\frac{A}{x+ 3}+ \frac{B}{x- 3}= \frac{A(x- 3)}{(x-3)(x+ 3)}+ \frac{B(x+ 3)}{(x- 3)(x+ 3)}= \frac{Ax- 3A+ Bx+ 3B}{(x- 3)(x+ 3)}= \frac{(A+ B)x- 3A+ 3B}{(x- 3)(x+ 3)}$.
As far as "Calculus II" is concerned, t

Sounds good.
 

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