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lcm(9, 6) = 18

So our answer is 18 sausages and 18 rolls. :))

Here's a video if you need to learn how to find the lcm: https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-factors-and-multiples/cc-6th-lcm/v/least-common-multiple-exercise 

=^._.^=

\(X * 2 * 230 = X * 10 - 10\)

solving for X gives

\(X = -10/450 = -0,02222222...\)

5) (n, k) are positive integers, solve n(n + 2) = k^2
6) (n, k) are positive integers, solve n(n + 4) = k^2

Hello Guest!

\(\{n,k\}\subset \mathbb Z\)

\(n(n+2)=k^2\)

             \(k\in\{ \) 4, 9, 16, 25, 36, 49, 64, 81, 100, 121   \(\}\)

             \(n\in\{ \) 1, 2,  3,   4,    5,  6,   7,   8,   9,   10,   11,   12\(\}\)

\(n(n+2)\in \{\)3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 168\(\}\)

No solution for positive n, but if n = -1

\(-1(-1+2)=i^2\)

\(n(n+5)=k^2\)

No solution for positive n, but if n = -4

\(-4(-4+5)=(2i)^2\)

  !

Food cost ----> $42.50

Tax Rate was 7.25%

And the tip was 15% of the food.

(7.25% x 42.50) + 42.50. This Equation will tell us how much they pay with tax (excluding the tip)

That answer is rounded off to $48.58

Now for the 15% tip.

(15% x 45.58) + 45.58 ≈ 52.42

Therefore, the Rodriguez family paid around $52.42.

Hope that helps :) !!!

🙃🙃👌

press shift (top left button) then the tan button

Hope that my diagram will answer your question.

Scale 1 : 1  There are a lot of different methods to calculate CM.

The top of a 15 metre high tower makes an angle of elevation of 60º with the bottom of an electric pole and angle of elevation of 30º with the top of the pole. What is the height of the electric pole?(in metres)

Hello Guest!

The pole is \(15m\cdot tan( 30°)\) away from the tower. So

\(15m-x=15m\cdot tan^2 (30°)\\ x=15m\cdot(1-tan^2(30^°))\)

\(x=10m\)

The height of the electric pole is 10m.

  !

50000 (  1+  .03/4) ^(5 * 4)   ≈  58059.21  php

{ php   =   Philippine pesos  }

Max does get the correct answer....and BMA is a right triangle.....but I don't see how he arrives at the  conclusion that this triangle is isosceles with BA   =  MA

The only way that would be  possible is that if CM  were a diameter  (which it isn't)

Am I misssing something,  jugoslav  ????

Try these coordinates:   C (0, 0)        M (7, 7)                                                                                                                 

 CM = √98 = 7√2  

We want to  first  assume that

ax + 3    =  x   -5       when  x  =  2    so

a(2) + 3  =   2 - 5

a(2)  =  -3 - 3

a(2)  =  -6

a  =  -3

Likewise.....we want to assume that

x-  5 =    2x  - b  when  x  =  -2

-2  -5  = 2(-2) - b

-7  = -4   -   b

b =  -4 +  7

b =  3 

a + b   = 0

See the graph here :  https://www.desmos.com/calculator/zh7glvlbum

Rate * Time  = Distance

10  (1)   +  24 (1)   =      34  miles

Guest.....why don't you make an account  ????

Your  answers are  very  good and very thorough   !!!!

Per jugoslav's diagram.....Let N be the intersection of the  chords BA and MC

Since  BCA is  bisected, we  have the following relationship

BC /CA  = BN /AN

6/8 =  BN /AN  = 3/4

So   BN  =  (3/7) 10  =  30/7

And  AN  = (4/7)10  = 40/7

And  AN * BN =  1200 / 49

Without re-inventing the  wheel  the length of the  angle  bisector is  given by

BC * CA sqrt (2)  /  ( BC + CA)  =   *8*6 sqrt (2)  / ( 8 + 6) =  48sqrt (2) / 14  =  24sqrt (2)  /  7  = CN

So

AN * BN  =   CN  *   NM

1200/49  = 24sqrt (2) / 7  * NM

1200/ 49  *   7  / (24 sqrt (2))  =  NM =   25sqrt (2)  / 7   

Therefore    CN +  MN  =    CM  =   (25 + 24) sqrt (2)  / 7  =    49sqrt (2)  /7  =   7sqrt (2) 

Here's the proof of the length of the  angle bisector  :

12  + 15  -  25  =      27  - 25 =    2 are wearing both

12 - 2   = 10   wear white  shirts  only

15  - 2    =  13    wear  black shoes only

2 + 10+  13  =  25

The  %   wearing both is          2  / 25    =    8 / 100  =    8%

The third equation is the  correct one

The first one  divisible  by  4  after  2018  is 2020

The last one  divisible  by 4   before  3018  is 3016

So.....the  number of  years   divisible  by  4   =    ( 3016 - 2020)  / 4   + 1     = 249 +  1  = 250

However......years  ending in  00  that are not multiples of 400 are not leap years

So

2100, 2200, 2300, 2500, 2600,2700, 2900 and 3000 are  not leap years =  8

So....the number of  leap years  between 2018  and 3018   =  250   -  8   =   242

There are three given points for this particular polynomial. Therefore, the minimum degree of this mystery polynomial must be of degree 2. The polynomial could be of a higher degree, but there is no need to consider such an option.

Since the polynomial is of degree 2, this polynomial is a quadratic. Quadratics are of the form \(f(x) = ax^2+bx+c\). Let's use the information given to use to determine the equation. Normally, this would be a nasty three-variable system of linear equations, but we can avoid this if we recognize that the y-intercept is already given.

Since the y-intercept is already given at the point (0, 5), \(c=5\). Unfortunately, we cannot take any more shortcuts beyond this point.

\(f(6) = a*6^2 + b*6 + c\\ 36a + 6b + 5 = 12\\ \fbox{1}\; 36a + 6b = 7\\ f(11) = a*11^2 + b*11 + c\\ 121a + 11b + 5 = -12\\ \fbox{2}\;121a + 11b=-17\)

Now, let's solve this system of two variables. Unfortunately, it is already clear that this system will be an algebraic nightmare, but we will manage and perservere. I will use the elimination method and eliminate b first as that is probably the simplest one to do at this point.

\(\quad 6*\fbox{2}\quad\quad 726a+66b=-102\\ -11*\fbox{1}\; -396a-66b=-\;\;77\\ \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\\ \quad\quad\quad\quad\quad\quad 330a \quad\quad\; =-179\\ a=-\frac{179}{330}\)

Ok, let's now solve for b. Neither equation looks as if there will be an easy substitution. I chose equation 1, but either one will do.

\(\fbox{1}\;36a+6b=7\\ 36*-\frac{179}{330}+6b=7\\ -6444+1980b=2310\\ 1980b=8754\\ b=\frac{8754}{1980}=\frac{1459}{330}\)

Therefore, the equation of this polynomial is as follows: \(f(x) = -\frac{179}{330}x^2+\frac{1459}{330}x+5\). Yikes! This one is awful to do by hand.

-3x  +  2y  - 6z   =  6

5x  +  7y  - 5z   =   6

 x +  4y   -   2z    =  8

Lookig at the bottom equation    ....mutiply  it through  by   3

3x  + 12y  - 6z  =  24             add this to the  first equation

14y  - 12z  = 30     divide through  by 2   ⇒  7y - 6z  = 15      (a)

Then, mutiply it through by    -5

-5x  - 20y  + 10z  =  -40         add this to the second equation

-13y  + 5z   =  -34       (b)

Mutiply (a)  through  by   5  and  (b)  through by  6

  35y  - 30z  = 75

-78y  + 30z  = -204          add these

- 43 y  = -129

y=  -129 / - 43  =   3

Using (a)    7(3)  - 6z = 15

                   21 -  6z  =  15

                    21 -  15  = 6z

                         6  = 6z

                          z    = 1 

And

x   + 4y  -2z  = 8

x + 4(3)   -2(1)  =  8

x  + 12 -2  = 8

x  + 10  = 8

x =  8 -10   =  -2

So   { x, y , z }  = { -2, 3 ,  1 }

omg haha thx so much!! that made me feel a lot better! :D

https://web2.0calc.com/questions/question-help_7   (Very clever!!!) Bravo, MaxWong!!!

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BC => a = 6         AC => b = 8           AB => c = 10

∠A = tan^-1(6 / 8)             

∠ANC = 180 - (45 + ∠A)

By using the Law of Sines we can find side AN of triangle ACN.

AN ≈ 5.714           BN = 10 - AN ≈ 4.286

CN can be calculated by using this formula: CN = sqrt [ab(a + b + c)(a + b - c)] / a + b

CN ≈ 4.849 

By using the Intersecting chords theorem we can calculate the length of CM.

MN * CN = AN * BN             MN ≈ 5.05

CM = CN + MN ≈ 9.899  

x +  y  =10    ⇒   y  =10 - x

So

x^2 + y^2  =   56 +  xy

x^2  +  (10 - x)^2  =  56  + x ( 10- x)

x^2  + x^2    -20x  + 100  =  56  + 10x  - x^2

3x^2  - 30x  + 44  =  0

Q Formula

x  =   30  ±  sqrt  [ 30^2  -  4 * 3 * 44 ]                 30 ± sqrt (372)

        ___________________________  =          ___________     =

                     2 *  3                                                      6

30 ±  2sqrt (93)

____________

         6

So       x  =   one  of these  solutions  and  y = the other

Actually, I bet I can!

\(prn+m\\ -8*4*2-32\\ -64-32\\ -96\)

Here you go!

This is a system of equations. Let's try and solve it.

While it is possible to solve for x or y and then do direct substitution into the remaining equation, I noticed a possible shortcut that is worth knowing for future algebraic problems like this one. This may not save too much time for this particular problem, but I figured it would be to your benefit if I showed this strange substitution to you.

\(\fbox{1}\; x + y = 10\\ (x+y)^2 = 10^2\\ x^2 + 2xy + y^2 = 100\\ {\color{red}x^2 + y^2} = 100 - 2xy \)

I squared equation 1 and rearranged a few terms. To the untrained eye, this may seem to complicate matters unnecessarily, but I am setting up a creative substitution.

\(\fbox{2}\; {\color{red} x^2 + y^2} = 56 + xy\\ 100 - 2xy = 56 + xy\\ 3xy = 44\\ y = \frac{44}{3x}\)

Let's use this substitution in equation 1 and solve it.

\(\fbox{1}\; x+y = 10\\ x+\frac{44}{3x}=10\\ 3x^2+44=30x\\ 3x^2-30x+44=0\)

This quadratic is neither easy nor factorable. I will result to the quadratic formula for this particular question.

\(x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a};a=3, b=-30, c=44\\ x_{1,2}=\frac{30\pm\sqrt{900-528}}{6}\\ x_{1,2}=\frac{30\pm\sqrt{372}}{6}\\ x_{1,2}=\frac{30\pm2\sqrt{93}}{6}\\ x_1=\frac{15-\sqrt{93}}{3}\text{ and }x_2=\frac{15+\sqrt{93}}{3}\)

Let's find the corresponding y-values by substituting these values for x into \(y=\frac{44}{3x}\).

There are two coordinates of intersections. One is at \(\left(\frac{15-\sqrt{93}}{3},\frac{44}{15-\sqrt{93}} \right)\), and the other is at \(\left(\frac{15+\sqrt{93}}{3}, \frac{44}{15+\sqrt{93}} \right)\)

.28 = 28 /100  =   7/25

7/25 =  1/n + 1/n^2   multiply through  by n^2

(7/25)n^2   =  n +  1            mutiply through by 25

7n^2   =  25n +  25

7n^2  - 25n  - 25  =  0

Q Formula

n =    25   -  sqrt  ( 25^2  +  4 * 7 * 25)              25   -  sqrt (25 [ 25 + 28 ] )

       ___________________________  =       ______________________   =

                    2 * 7                                                         14

25  -  5  sqrt (53)

_______________

          14