Melody

A grocery clerk stacks rows of fruit cans for a display. The number of cans left to stack is 156-8r where r is the number of rows she has stacked. What does each part of the expression represent?

156 would be how many cans there were to start with.

Every time you display a row it takes 8 cans so                              

the number of rows you have already stacked is r 

The number of cans already stacked is 8r

The number of cans left to stack is   156-8r

Draw a pic of the stacking cans if you are still confused.     Put 8 cans in each row!

You can do the second one yourself.  :)

Why does -2.5 - 0.5 equal - 3, but 2.5 - 0.5 equal 2?

Good question, it is important that you understand this :)

I am showing you on a number line.

When you are subtracting one numbers from another, you always start at the first number,  

and then move left towards the negative end of the numberline how ever many units you want to subtract.

So 

For      -2.5 - 0.5   start at -2.5 then go in the negative direction (left) 0.5 units and you will come to -3

For     2.5 - 0.5  start at 2.5 then go in the negative direction (left) 0.5 units and you will come to +2

Find the domain of $\frac{x^2 + 10x + 21}{x^2 + 4x - 21}$. (Express your answer using interval notation.)

\(\frac{x^2 + 10x + 21}{x^2 + 4x - 21}\\ =\frac{(x+3)(x+7)}{(x-3)(x+7)}\\ x\ne-7 \qquad x\ne3\\ Domain\quad(-\infty,-7)\cup (-7,3) \cup(3,\infty)\)

I think this is interval notation..?...

Factor and express the answer as a quotient with positive exponents.

x^(-2)-4x^(-5)+3x^(-8)

\(x^{-2}-4x^{-5}+3x^{-8}\\ =x^{-8}(x^6-4x^3+3)\\ =\frac{x^6-4x^3+3}{x^8}\)

(cotx/sinx-tan^2x- tanx/cos+2x/cos^2x)dx

NOTE I have added an x after cos as I think that you forgot it.  :)

\(\int (\frac{cotx}{sinx}-tan^2x-\frac{ tanx}{cosx}+\frac{2x}{cos^2x})dx\\ =\int \frac{cotx}{sinx}dx-\int tan^2x dx-\int\frac{ tanx}{cosx} dx+\int\frac{2x}{cos^2x}dx\\ =\int \frac{cotx}{sinx}dx-\int tan^2x dx-\int\frac{ tanx}{cosx} dx+\int\frac{2x}{cos^2x}dx\\~\\ -----\\ \int \frac{cotx}{sinx}dx=\int \frac{cosx}{sin^2x}dx=\int cosx(sinx)^{-2}dx=-(sinx)^{-1}+c_1=-cosecx+c_1\\ \int tan^2x dx=\int \frac{sin^2x}{cos^2x} dx=\int\;(sec^2x-1)dx=tanx-x+c_2\\ \int\frac{ tanx}{cosx} dx=\int\frac{ sinx}{cos^2x} dx=(cosx)^{-1}+c_3=secx+c_3\\~\\ \text{So far I have }\\ =\:-cosecx-(tanx-x)-(secx)+\int\frac{2x}{cos^2x}dx\\ =\:-cosecx-tanx+x-secx+\int\frac{2x}{cos^2x}dx \)

\(=\:-cosecx-tanx+x-secx+\int\frac{2x}{cos^2x}dx\\ consider\;\;\;\int\frac{2x}{cos^2x}dx\\ =\int(2x(sec^2x)dx\\ \text{Solve using integration by parts}\\ u=2x \qquad v'=sec^2x\\ u'=2 \qquad v=tanx\\ \int\;uv'dx=uv-\int(vu')dx\\ =2xtanx-\int(2tanx)dx\\ =2xtanx-2\int(\frac{sinx}{cosx})dx\\ =2xtanx+2ln(cosx))+c_4\\~\\ \text{So my final answer is}\\ =\:-cosecx-tanx+x-secx+2xtanx+2ln(cosx)+c\\ \)

Please include your question in the body of your post .. not just in the title :/

x-y=5      (1a)          becomes       x=5+y  (1b)

-4x+4y=20  (2a)     becomes         -x+y=5     becomes      x=y-5 (2b)

Put these together and we get

5+y=y-5

That is never going to happen :(

So there are no solutions.   Yes it is very sad :((

What so funny ?? 

58 is 58.0 and you can add as many more 0s as you like.

When you subrtract or add then you have to line up the points :)

So the 8 has to go above the 3  

The 8 and the 3 are both in the units column :)

Hi Juriemagic,

It is good to see you again :)

what you have done is correct but you would neer leave it like that :)

S=P(1+rt) 

\(S=P(1+rt) \\ t=(\frac{S}{P}-1)\div r \\ t=(\frac{S}{P}-1)\times \frac{1}{r} \\ t= \frac{1}{r}(\frac{S}{P}-1)\\ t= \frac{S}{Pr}-\frac{1}{r}\\ \)

Yea, I like my answer better ;)