Melody

Yes, it is cool :)

Didn't you like my explanation Chris?    

Actually, I do like yours, but mine is simpler because it has no algebra.

I get 5/14 which is the same as our guest.

Ruan began painting a room at 9:00 AM. Leonardo, who can paint twice as fast as Ruan, started helping Ryan at 9:20 AM, and they worked together until the room was fully painted at 10:00am. What fraction of the room had been painted by 9:30 AM? Express your answer as a common fraction in lowest terms and clearly show/explain your reasoning.

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Here is my logic   

Let Ruan paint one unit of area every minute.

then Leonado is twice as fast so he will paint 2 units of area ever minute.

In the whole hour Ruan will paint  60 units of area

Leonado only paints for 40 minutes so he will paint a total of 80 units of area.

So in the hour they paint 60+80 = 140 units of area and this is the whole room.

Now how much is painted by 9:30am

Ruan has painted for 30 minutes so he has painted  30 units of area.

Leonado has painted for 10 minutes so he has painted 20 units of area.

So in the first half hour  30+20 = 50 units of area is painted.

So that is    \(\frac{50}{140}=\frac{5}{14}\)       of the room is painted by 9:30

"both came out wrong" is neither nice nor mean. It is certainly not polite though.

Plus, I had not idea what you were talking about. At least you have now explained better why you think it was wrong.

Maybe it is your program that has the wrong answer, did you ever consider that?

I'll take (a) look at a for you.

P = 9+9+13+13+7+7+   the arc

= 58mm + the arc

The arc is from a circle with radius9 and subtended from an angle of 44 degrees

so

 \(length \;\;of \;\;arc\;=\frac{44}{360}*2*\pi*9\\ length \;\;of \;\;arc\;=\frac{22}{10}*\pi\\ length \;\;of \;\;arc\;=2.2\pi \approx 6.9mm \)

\(Perimeter=(58+2.2\pi)mm \approx 64.9mm\)

I think CPhill misinterpreted what a composite figure is but he did all the hard bits, you should have been able to figure out why his answer was not correct yourself.  Did you even try to?

Angie, it is good that you interacted with Coldplay and you gave her a point as well, that is great!     

It would have been nice if you had also slipped in a 'Thank you'.  Thank yous always mean a lot.   

Still what you did was really good.   

CPhill tried to help you, you said he was wrong but you did not why you believe this to be true.

Go back and interact with CPhill properly!

Why do you say these are wrong Cassie?

You need to give a reason.

For example.

"Thanks for you time CPhill but your answers do not match with the answers in my text book.

The answers in my book are ......   "

ALWAYS THANK PEOPLE, EVEN IF IT IS JUST FOR THEIR TIME.   (which they give to your for free)

The polynomial f(x) has degree 3. If f(-1)=15, f(0)=0, f(1)=-5, and f(2)=12, then what are x-intercepts of the graph f?

\(f(x)=ax^3+bx^2+cx+d\)

\(f(0)=0\qquad so \qquad d=0\\ f(x)=ax^3+bx^2+cx\)

\(also\\ f(-1)=15,\;\;  f(1)=-5, \;\;and \;\; f(2)=12\)

so

\(f(-1)=-a+b-c=15\qquad(1)\\ f(1)=a+b+c = -5\qquad\quad(2)\\ f(2)=8a+4b+2c=12\qquad(3)\\ (1)+(2) 2b=10\\ b=5\\\)

Now we have

\(-a+5-c=15\qquad(1)\\ a+c=-10 \qquad \qquad (1b)\\ \\~\\ 8a+20+2c=12\qquad(3)\\ 4a+10+c=6 \\ 4a+c=-4 \qquad \qquad(3b)\\ \\~\\ (3b)-(1b)\\ 3a=6\\ a=2\\ c=-12 \)

\(f(x)=ax^3+bx^2+cx\\ f(x)=2x^3+5x^2-12x\\ f(x)=x(2x^2+5x-12)\\ f(x)=x(2x^2+8x-3x-12)\\ f(x)=x[2x(x+4)-3(x+4)]\\ f(x)=x[(2x-3)(x+4)]\\ \text{Roots are x = 0, 1.5 and -4}\)

Sounds like a very good guess to me. :)