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r [  c1     c2                      c3                         c4     ....                          ]

           r*c1                 r * [  r*c1 + c2]      r [  r * [ r*c1 + c2] + c3 ]        ]

     ____________________________________________________

     c1    r*c1 + c2     r * [ r*c1 + c2] + c3     r [  [ r*c1 + c2] + c3 ] + c4       .....etc

OK....using your formula from last week...we have

1/3 + 1/4  =  1/ Total Time Working Together

4/12  + 3/12 =  1 / T

7 /12  = 1 / T       [ remember that we can write ]

12 / 7  =  T / 1

12/ 7  = T     

T = 12/7 hrs ≈     1.7 hrs  =    1 + .7(60)  =   1 hr 42 min

x - y - z = 0

2x + 3y + z = 17

z + 2y = 7

No set way to do this GM.....but here is an approach....the idea is to get to two equations and two unknowns

Using the 3rd equation.....we can write    z = 7 - 2y

Sub this into the first two equations for z

x - y - [ 7 - 2y ]  =  0     ⇒    x - y - 7 + 2y = 0   ⇒    x + y    = 7     (4)

2x + 3y + [ 7 - 2y ] = 17 ⇒   2x + y  + 7 = 17 ⇒    2x + y = 10      (5)

We can rewrite (4)  as  y = 7 - x

Sub this into (5)

2x + ( 7 - x) = 0

x + 7 = 10

x = 3

So...y =  7 - x =   7 - 3  = 4

And

z = 7 - 2(y)   =  7 - 2(4)  = - 1

So   { x, y, z }      =      { 3, 4 , -1 }

a)  We can make up whatever we want

For the first....a possibility is    ( x - 3)^5 (x - 2)^2

[ We have repeated real zeroes, but the problem doesn't say that every zero has to be unique ]

The graph of this is here :   https://www.desmos.com/calculator/td0qq1sakf

b)  We can have

(x  - 3) ( x + 3) ( x - 4) ( x + 4) (x + 6) ( x^2 + 1)

The graph of this is here : https://www.desmos.com/calculator/0f8x7e1irp

c) We can have

(x^2 + 1) ( x^2 + 3)

Graph : https://www.desmos.com/calculator/sqj7sk0rcf

d) We can have

(x^2 + 1) (x + 3) ( x - 3) ( x + 2) ( x - 2)

Graph : https://www.desmos.com/calculator/nyabj6mmtz

I hope these links work, GM !!!

 (x-4)(x-8) = 12

Expand the left side

x^2 - 12x + 32 =  12        subtract 12 from both sides

x^2 - 12x + 20  =   0

Factor

(x  - 10 ) ( x - 2) = 0

Setting each factor to 0  and solving for x produces

x = 10    or  x  = 2

Thanks....I missed the "acute" part

The triangle  will be obtuse if  this is true

missing side > sqrt ( 15^2 + 8^2) 

missing side > sqrt (289)

missing side > 17

And when x = 17...we have a right triangle

So.... the triangle will be acute when  x =    8, 9, 10, 11, 12, 13, 14, 15, 16

So....9 values are possible for an acute triangle

Okay, think I got it. Just double checking:

     r |     c₁       c₂   …   c_n

                           r•x₁  …  r•x_n

                x₁         x₂   …   x_n

Did I get it? (Pls say yes, pls say yes...) XD Thanks a ton man! You probably save my grade from going to an F!

ignore the dollar signs those are for LateX.

It's really pretty simple

Since we   are   testing if x - 4  is a factor....we actually want to see if 4 is a root....this is what we divide by

The coefficients (and constant term ) of the polynomial are then written.....be careful....any missing power is represented by 0

So we have

4 [    1pi          -5pi          -86pi          360pi     ]              bring down the first term

      ___________________________________  

        1pi                                                                

Multiply this by the divisor....and put this result  (4pi) under the second term....then add

So we have

4 [ 1 pi     - 5pi       - 86pi       360pi   ]

                  4pi

     ____________________________       

     1pi       -1pi   

Multiply  4 (-1pi)  and put the result  (-4pi)under the third term.....add again 

4 [ 1 pi     - 5pi       - 86pi       360pi   ]

                  4pi        -4pi      

     ____________________________       

     1pi       -1pi      - 90pi 

Multiply  4(-90i)  and put the result  (-360 pi) under the last term....add again and we get 0

4 [ 1 pi     - 5pi       - 86pi       360pi   ]

                  4pi        -4pi        -360pi      

     ____________________________       

     1pi       -1pi      - 90pi          0

The remaining polynomial  is  one degree less than the one we started with   ..so...    degree 2

So we have

1pi x^2   -  pi x  - 90 pi           factoring out pi, we get

pi [ x^2 - x  - 90 ]   =  the remaining polynomial

And that's it!!!

Hope that helped

\((2y-5)^2 + (y-15)^2 = 130\\ 4y^2 - 20y+25 + y^2 - 30y+225 = 130\\ 5y^2 - 50y +120=0\\ y^2 - 10y+24=0\\ (y-6)(y-4) = 0\\ y=6,~4\\ x=12,~8\\ max(x) = 12 \)

Thanks, Rom, but I have to do it in stupid synthetic division or I get the entire question wrong.

Thanks, CPhill! So, essentially, when you use synthetic division, remove the variable, execute long division on each coefficient by the factor you are testing, and reinsert the variable with respect to each part's powers? Or am I still wrong? Because I'm still quite lost with synthetic division. I mean, I would use long division, but no, we have to use synthetic division because it makes that much of a difference in the long term.

\(\text{what you think you have to prove can't be right because}\\ 2^6+1 = 65 = 5 \cdot 13 \text{ isn't prime, and}\\ 2^9-1 = 511 = 7\cdot 73 \text{ isn't prime either}\\ \text{so that can't be a general rule regarding the primality of powers of two plus or minus 1}\)

\(\text{you just need to prove that the two numbers are coprime, i.e. have no common factors}\)

When (0,0) is reflected about the vertical line x = 1, the new point is (2,0)

When this point is reflected across y = 2, the final point is   (2, 4)

When P is reflected across x, P'  =  (2, - 1)

When R is reflected across x, R' = (12, -15)

So M' =  [ (2 + 12) / 2, (-1 - 15)/2 ]   =  [14/2, -16/2 ] = (7, -8)

So...the sum of the coordinates of M'  =   7 + - 8    = -1

Not that difficult, really...

4 [    1pi          -5pi          -86pi          360pi     ]

                        4pi           -4pi          -360pi                 

       ___________________________________

        1pi          -pi           -90pi             0

So....the remaining polynomial, p(x) ,  is  

p(x)  = pi [ x^2 - 1x   - 90 ]

\(\text{one way of doing this is to note that the rotation matrix for a 90 degree clockwise rotation is}\\ R=\begin{pmatrix}0 &1\\-1 &0 \end{pmatrix}\)

\(R.(x,y) = (y, -x)\\ \text{and this is simple to apply to your points}\\ \text{for example}\\ (-4,1) \to (1,4)\\ \text{I leave rotating the two other points to you}\)

Do you have a diagram for the previous problems... 

1. I'll give a solution just for this one: Take the reciprocal like the user above had said. 

1. \(\frac{2p}{4p^2-1}\times \frac{6p+3}{6p^3} \)

2. We realize that \(\frac{6p+3}{6p^3}=\frac{2p+1}{2p^3}.\)

3. We get: \(\frac{2p\left(2p+1\right)}{\left(4p^2-1\right)\times \:2p^3}\).

4. And. \(\frac{p\left(2p+1\right)}{\left(4p^2-1\right)p^3}\), so \(\frac{2p+1}{p^2\left(4p^2-1\right)}\).

5. Factor: \(4p^2-1:\quad \left(2p+1\right)\left(2p-1\right)\)

6. Therefore, \(\frac{1}{p^2\left(2p-1\right)}\) , consequently, \(p^2\left(2p-1\right):\quad 2p^3-p^2.\)

Thus, the final answer is \(\boxed{\frac{1}{2p^3-p^2}}.\)

Take the reciprocal: 1/3 + 1/4= 7/12. So, 12/7=1 5/7 hours. 

Simplify 

2 log(x) + 3 log(y) - 4 log(x y) = 0

Combine logarithms.
2 log(x) + 3 log(y) - 4 log(x y) =   

log(x^2) + log(1/(x^4 y^4)) + log(y^3)

log((x^2 y^3)/(x^4 y^4)) = log(1/(x^2 y)):
log(1/(x^2 y)) 

1/4  +   1/3 =1/T

1/T = 7/12        Take the reciprocal of both sides

T =12/7 =1 5/7 - hours =~ 1 hour and 43 minutes for both pumps to fill the tank.