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7/16 x  = pencils where  x = total number of writing instruments

7/ 16 x  + 1  = 5/12 (x+4)

21/48  x + 1 = 20/48 x + 80/48

1/48 x = 32/48

x = 32    pencils+ pens originally  

7/16 (32) = 14  pencils originally   ... and you found one more    = 15

it's 22?

it's C

oh it's D

It's C?

I have a pouch with pens and pencils in it. At the end of math class, the ratio of pens to pencils is 9:7. While walking from math class to photography class, I found three pens and one pencil. After I put them in my pouch, the ratio of pens to pencils was 7:5. How many pencils did I have at that point?

Call x the number of pens tha you had at first....and y the number of pencils you had at first

So....we now that

x / y =   9/7      which implies that  x = (9/7)y       (1)

Afteryou found three pens and one pencils we know that

[ x + 3] / [ y + 1]  =  7/5        cross-multiply

5[ x + 3 ] =  7 [ y + 1]

5x + 15  =   7y + 7           sub (1) in for x

5(9/7)y + 15 = 7y + 7      simplify

(45/7)y + 15 = 7y + 7

15 - 7  =   7y - (45/7)y

8 =  (49/7)y - (45/7)y

8 = (4/7)y 

6 (7/4) = y =   14   = no. of pencils initially

And you added one...so....now you have 15

Check

You had   (9/7)y = (9/7)(14) =  (9 )(14/7) = 9 * 2 = 18 pens originally

So    18/14  =  9/7

And at the end you had    [ 18 + 3] / [ 14 + 1]  =   21 / 15   =  7 / 5

LOL!!!!.....yeah...it's hard to solve without doing that !!!!!

oh wow, seems like i forgot to differentiate it....

thanks!

I see where I messed up... I though 12/2 = 12! 

h ' (t)  =   12t^2 - 59t + 72

So

h ' (3) =   12(3)^2 - 59(3) + 72   =  108 - 177 + 72  =   180 - 177   =  3 cm / hr

Is there a diagram / picture for this problem?

I forgot what  \( |_x_|\) means...

I think you messed up on the LaTeX in C and D. I cant read it...

sorry,

- Pm

If you want an actual explanation, feel free to ask for one.

b) Just plug in numbers 1, 2, and 3.

1*2*3 = 6. 6/5 = 1 R1.

Remainder \(\boxed{1}\).

- PM

When x^4+6x^3-ax^2-45x-15 is divided by x^2-x-6 the remainder is 3. What is a?

Haven't done one like this before......!!!!

                        x^2  + 7x+ (13 - a)

x^2 - x  - 6   [   x^4 + 6x^3 - ax^2    -   45x     -    15 ]

                        x^4 - x^3 -  6x^2

                       _____________________________

                               7x^3 + (6 - a)x^2 - 45x

                               7x^3 -     7x^2      -42x

                               ___________________________ 

                                          (13 - a)x^2 - 3x          - 15

                                          (13 - a)x^2 - (13 -a)x - 6(13 - a) 

                                         ___________________________

                                                              (10 - a)x  - 15 + 6(13 - a)

So

-15 + 6(13 - a) = 3

-15 + 78 - 6a = 3

63 - 6a = 3

-6a = -60

 a = 10

Check here to see that this is true : 

https://www.wolframalpha.com/input/?i=%5B+x%5E4+%2B+6x%5E3+-+10x%5E2+-+45x+%C2%A0-%C2%A0+15%5D+%2F+%5B+x%5E2+-+x+-6%5D

It is an ellispe because

putting it into conic form we have 

\((x^2-6x+9)+2(y^2-10y+25)=12\)

\((x-3)^2+2(y-5)^2=12\)

\( \frac{(x-3)^2}{12}+\frac{(y-5)^2}{6}=1\)

you can also put it into desmos to see the graph of it

For number 4 the answer is not 54? so what will it be? CPhill

Hahaha love it, thanks!

When x^3+30x-c is divided by x-5 the remainder is 5. What is c?

We can use some synthetic division to solve this

5   [  1   0      30       - c   ]

             5      25      275

        ________________

        1    5    55       275 - c

So

275 - c = 5     implies that  c = 270

\(\text{just use the quadratic formula}\\ r_1,r_2 = \dfrac{-7\pm\sqrt{49-8c}}{4}\\ \dfrac{1}{r_1}+\dfrac{1}{r_2} = \dfrac{4}{-7+\sqrt{49-8c}}+ \dfrac{4}{-7-\sqrt{49-8c}}=\\ \dfrac{-56}{49-49+8c} = \dfrac{-56}{-8c}= \dfrac{7}{c}\)

Let the external point = C

Let the tangent points be A and B

Let the circle center = 0

Draw OC, OA and OB

And we have two triangles  OCA  and OCB

Now OA and OB are radii drawn to tangent points......so they meet the tangents at right angles

And OA = OB since they are radii

And CO = CO 

So....CO is the hypotenuse of two right triangles

And OA and OB are equal legs

So...by Hypotenuse - Leg, the other two legs of both triangles - CA and CB - are equal

But.....CA and CB are the tangents drawn from C

So.....they are equal

1)

 The only obvious solutions are:

x = 0   and   y = 0, and:

x = 2   and   y = 2

2) You have 4 letters a, b, c, d instead of just 3, a, b, c !!!

Note that  we can write this in a slightly different manner

1 + (-1/3 + 1/3) + (-1/5 + 1/5) + (-1/7 + 1/7) +......+ (-1/17 + 1/17) - 1/19    =

1 - 1/19  =

18 /19    

\(\dfrac{10}{18}\dfrac{9}{17} + \dfrac{8}{18}\dfrac{7}{17} = \dfrac{73}{153}\)