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8 (X+2) + 7 (X-3) - 5 (X-4)  =

8X + 16 + 7X - 21 -5X + 20      combine like terms

10X  + 15

OK. What you have to do is find the "Net Present Value" of ALL 15 positive cash flows discounted @ 10%. You should get $238.61. You would use this as PV, or present value. Then you have 15 terms as "n". You already have the interest rate of 10%. Then you solve for "Net Uniform Series" of payments. If you haven't made  a mistake, you should get $31.37 as your "NUS" payment.

If you have any problems, just let me know here. Good luck to you.

(1.496 x 10^11) / (5.2 x 10^-3)   we can write this as :

[1.496 / 5.2 ]   * [ 10^11 * 10^3  ]  ≈

.2877   *    10^14  ≈

2.877 * 10^13

yes

Just input it normally.......the calculator will handle all the basic operations with fractions

Is the question about " Cash Flow Analysis"? And that the initial investment is $80 and have positive cash flows of 30, 40, 50 for 5 periods each? Am I reading it right?.

2x^2 + 5x + b  = 0

If b is a positive integer.......it cannot be an integer larger than 3 because the discriminant  would be < 0

So......the only possible positive integer values for b are either 1, 2 or 3

If b =  1, the discriminant   =  √17 and this will  not  produce a rational  solution

If b = 2, the discriminant = √9  = 3    and this will produce a rational  solution

And if b = 3, the discriminant   = √1   = 1   and this too will produce a rational solution

So....2,3 produce rational roots for b....where b is a positive integer

And the sum of these  = 5

Uniform Series 

What does "A" represent?

Sorry....it should be   3/4 + √3  =   [ 3 + 4√3 ] / 4

CPhill: I think there is a small error in your last result.

sin pi/3 =   √3/2    so the square of this is 3/4

tan pi/3   = √3/1     

So  √3/4  + √3/1   =  [ √3 + 4√3] / 4   =  [5√3] / 4

Solve for x:
15 x^2+28 x+12 = 9 x^2-18 x-16

Subtract 9 x^2-18 x-16 from both sides:
6 x^2+46 x+28 = 0

The left hand side factors into a product with three terms:
2 (x+7) (3 x+2) = 0

Divide both sides by 2:
(x+7) (3 x+2) = 0

Split into two equations:
x+7 = 0 or 3 x+2 = 0

Subtract 7 from both sides:
x = -7 or 3 x+2 = 0

Subtract 2 from both sides:
x = -7 or 3 x = -2

Divide both sides by 3:
Answer: |  x = -7 or x = -2/3

9x^2 -18x - 16  = 0

15x^2 + 28x + 12 = 0     

Since both quadratics = 0, set them = to each other

9x^2 -18x - 16  = 15x^2 + 28x + 12       subtract 9x^2 from both sides, add 18x, 16 to both sides

6x^2 + 46x + 28  = 0     divide everything by 2

3x^2 + 23x + 14   = 0     factor

(3x + 2) (x + 7)   = 0     setting both factors to 0 and solving,  x = -2/3   or x = -7

However since we're trying to find where both functions = 0........the solution is x = -2/3

The second solution of x = -7 is another intersection point of both graphs

   sin² π/3 + tan π/3 what is the answer?  =3/4+sqrt(3)

Max:  Here is one I can teach to you,

you showed

If you want to make summation sign, do \sum_{lower limit}^{upper limit} (For infinity sign, do \infty)

Example: Latex code:\sum_{n=1}^{\infty}                \(\sum_{n=1}^{\infty} \)

This is how I would do it:

\displaystyle \sum_{lower limit}^{upper limit} (For infinity sign, do \infty)

\(\displaystyle \sum_{n=1}^{\infty} \)

Thanks Max and asinus, it is good for people to see how LaTex can be used.  :)

I had vaguely wondered about this one      180^{\circ}

I have always used  180^0  but yours is better max    

I will have to try and remember it :)

A=9B

A+B=180

so

9B+B=180

and you can finish it :)

Find the sum of all possible positive integer values of b such that the quadratic equation 2x^2 + 5x + b = 0  has rational roots.

This will have rational roots when

\(25-4*2*b\ge 0\\ 25-8b\ge0\\ -8b\ge-25\\ 8b\le25\\ b\le 3\frac{1}{8} \)

b has to be a positive integer - according to the question so

b can equal 1,2,or 3

1+2+3 = 6

Hi guests - I am not sure if you are the same guest or 2 different guests but I iwll take a look. 

This is a very poorly worded question. !!

fuel cost involved in idling your car for 5 minutes per day when the cost of gasoline is $3.75(per gallon) using formula c=19a-4 where c is the cost (I will assume this is in cents) and a is the size of the engine in liters. Determine the size of the engine in your car and find the cost of idling your car for 5 minutes a day. cost of gasoline is $3.75(per gallon) and engine size is 4.0 liters

C = 19a - 4

C is total cost - not the cost of per gallon

a=4

C=19*4 - 4

C = 76 -4

C = 72 cents

so yes guest - you interpreted the question the same as I did but really the question did not make sense.  :)

Hi Pro35hp and MaxWong,

I think that you will both find this interesting.   It ties in nicely with Alan's graph as well.     

If you do not understand then please ask questions :)

\(\frac{x^2 + x + 3}{2x^2 + x - 6} \ge 0\\ \frac{x^2 + x + 3}{2x^2 + 4x-3x - 6} \ge 0\\ \frac{x^2 + x + 3}{2x(x+2)-3(x+2)} \ge 0\\ \frac{x^2 + x + 3}{(2x-3)(x+2)} \ge 0\\~\\ \mbox{Now I will consider the numerator and put it equal y}\quad y=x^2+x+3\\ \mbox{The discriminant is }1^2-12=-11 \qquad \mbox{ This means there are no real roots}\\ \mbox{Since the leading coefficiant is positive that means that }\quad y=x^2+x+3 \quad \mbox{is a concave up parabola}\\ \mbox{This means that for all real values of x the numerator will be positive}\ so\\ \mbox{If the denominator is > or = zero the statement will be true.}\\ True\;\;when\;\;\\ (2x-3)(x+2)\ge0\\ \mbox{If I graph }y=(2x-3)(x+2) \mbox{then this will be true when y is on or above the x axis, that is when } y \ge0\\~\\ \mbox{This is a concave up parabola so it will be true at the two ends}\\ \mbox{roots are } x=3/2\;\;\;and\;\;\ x=-2\\ \mbox{So the original statement will be true when }x\le-2\quad and \;\;when\;\; x\ge 1.5 \)

If y=arctan x, show that dy/dx=1/1+x squared, using the inverse of trigonometric functions. 

I am not sure if this was the intended method - I always do these from scratch.

\(y=atan(x)\\ x=tan(y)\\ \frac{dx}{dy}=sec^2(y)\\ \frac{dy}{dx}=\frac{1}{sec^2(y)}\\ \frac{dy}{dx}=cos^2(y)\\~\\ Given\;\; tan(y)=\frac{x}{1}\;\;then\;\;\\ If \;\;opp=x\;\;and\;\;adj=1\;\;then\;\;hyp=\sqrt{1+x^2} \;\;so\\ cos^2(y)=(\frac{1}{\sqrt{1+x^2}})^2=\frac{1}{1+x^2}\\~\\ \frac{dy}{dx}=\frac{1}{1+x^2}\\\)

f(x+1)=x.f(x)       f(5)=72      f(2)=?

\(f(5)=4*f(4)\\ 72=4*f(4)\\ f(4)=18\\~\\ f(4)=3*f(3)\\ 18=3*f(3)\\ f(3)=6\\~\\ f(3)=2*f(2)\\ 6=2*f(2)\\ f(2)=3\)

Please excuse.
A LaTeX - exercise.

\(4\frac{3}{4}\)  +  \(\sqrt[3]{8}\)  =  \(\frac{16\ +\ 3}{4}+2 = \frac{19}{4}+ \frac{8}{4}=6\frac{3}{4} \)  
4\frac{3}{4}+\sqrt[3]{8}=\frac{16\ +\ 3}{4}+2 = \frac{19}{4}+ \frac{8}{4}=6\frac{3}{4}

\(\frac{16\ +\ 3}{4}+2 = \frac{19}{4}+ \frac{8}{4}=6\frac{3}{4} \)   

\frac{16\ +\ 3}{4}+2 = \frac{19}{4}+ \frac{8}{4}=6\frac{3}{4}

This graph might help: