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Should I be worried if I'm having a hard time with this first assignment?
Hi. I'm taking calculous 1 this semester. I haven't taken a math class in 8 years, and the last class I took was trig. I studied pretty hard and placed in calculous through the assessment test. The administrator said, she had never seen anyone score as high as I did on the placement test. But, this first assignment, which is a review of algebra and trig, seams considerably harder than the placement test, and I am struggling on many of the problems.
If someone could look at the review sheet, and give an opinion on wether or not a person struggling with this assignment should take calculous 1, or if maybe these problems are intentionally on the difficult side to force us to do a serious review.
Thanks.
Re: Should I be worried if I'm having a hard time with this first assignment?
Quote:
Originally Posted by
StudentMCCS 
Hi. I'm taking calculous 1 this semester. I haven't taken a math class in 8 years, and the last class I took was trig. I studied pretty hard and placed in calculous through the assessment test. The administrator said, she had never seen anyone score as high as I did on the placement test. But, this first assignment, which is a review of algebra and trig, seams considerably harder than the placement test, and I am struggling on many of the problems.
If someone could look at the review sheet, and give an opinion on wether or not a person struggling with this assignment should take calculous 1, or if maybe these problems are intentionally on the difficult side to force us to do a serious review.
Thanks.
It seemed pretty easy to me. Which questions exactly bother you?
Re: Should I be worried if I'm having a hard time with this first assignment?
Quote:
Originally Posted by
abhishekkgp It seemed pretty easy to me. Which questions exactly bother you?
Question number 4 is bothering me a little. Looking at my list of identities, and thinking about an approach, I'm not seeing the light yet. Probably something simple I fail to recognize.
Re: Should I be worried if I'm having a hard time with this first assignment?
Quote:
Originally Posted by
StudentMCCS Question number 4 is bothering me a little. Looking at my list of identities, and thinking about an approach, I'm not seeing the light yet. Probably something simple I fail to recognize.
i don't know what is meant by "rational expression" so i can't help. But if this is the question which you couldn't do then why are you worried?
Re: Should I be worried if I'm having a hard time with this first assignment?
Quote:
Originally Posted by
abhishekkgp i don't know what is meant by "rational expression" so i can't help. But if this is the question which you couldn't do then why are you worried?
I'm also having trouble with number 6.
(((A+rA-x)r-x)r-x)r-x=o
solve for x.
Simplified
(AR^3)-(2AXR^2)+(ARX^2)+(AR^4)-(AXR^3)+AX^2(R^2)-(XR^3)+(2X^2)(R^2)-(XR^3)-x=0
Obviously this is leading me nowhere.
maybe start by saying (((A+RA+X)R-x)R-X)R=x
Re: Should I be worried if I'm having a hard time with this first assignment?
Quote:
Originally Posted by
StudentMCCS I'm also having trouble with number 6.
(((A+rA-x)r-x)r-x)r-x=o
solve for x.
Simplified
(AR^3)-(2AXR^2)+(ARX^2)+(AR^4)-(AXR^3)+AX^2(R^2)-(XR^3)+(2X^2)(R^2)-(XR^3)-x=0
Obviously this is leading me nowhere.
maybe start by saying (((A+RA+X)R-x)R-X)R=x
![(((A+rA-x)r-x)r-x)r-x=0 \Rightarrow (((A(r+1)-x)r-x)r-x)r-x=0 \Rightarrow ((Ar(r+1)-rx-x)r-x)r-x=0 \Rightarrow ((Ar(r+1)-x(r+1))r-x)r-x=0 \Rightarrow (((Ar-x)(r+1))r-x)r-x=0 \Rightarrow r^2 (Ar-x)(r+1)-xr-x=0 \Rightarrow r^2 (Ar-x)(r+1)-(r+1)x \Rightarrow (r+1)[r^2 (Ar-x)-x]=0](http://latex.codecogs.com/png.latex?(((A+rA-x)r-x)r-x)r-x=0 \Rightarrow (((A(r+1)-x)r-x)r-x)r-x=0 \Rightarrow ((Ar(r+1)-rx-x)r-x)r-x=0 \Rightarrow ((Ar(r+1)-x(r+1))r-x)r-x=0 \Rightarrow (((Ar-x)(r+1))r-x)r-x=0 \Rightarrow r^2 (Ar-x)(r+1)-xr-x=0 \Rightarrow r^2 (Ar-x)(r+1)-(r+1)x \Rightarrow (r+1)[r^2 (Ar-x)-x]=0)
Re: Should I be worried if I'm having a hard time with this first assignment?
For exercice 4 I just think a rational expression is an expression of the form 
so which have a numerator and a denominator. In this case you can just use some well known trig formulas:
![\sec(x)\cdot \tan(x)-\sqrt{2}\sec^2(x)=\sec(x)\cdot[\tan(x)-\sqrt{2}\sec(x)]](http://latex.codecogs.com/png.latex?\sec(x)\cdot \tan(x)-\sqrt{2}\sec^2(x)=\sec(x)\cdot[\tan(x)-\sqrt{2}\sec(x)])
![=\frac{1}{\cos(x)}\cdot \left[\frac{\sin(x)}{\cos(x)}-\frac{\sqrt{2}}{\cos(x)}\right]=\frac{\sin(x)-\sqrt{2}}{\cos^2(x)}](http://latex.codecogs.com/png.latex?=\frac{1}{\cos(x)}\cdot \left[\frac{\sin(x)}{\cos(x)}-\frac{\sqrt{2}}{\cos(x)}\right]=\frac{\sin(x)-\sqrt{2}}{\cos^2(x)})
Which is in my opinion a rational expression.
