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4/3 divided 4/3   

Anything divided by itself is 1.  Except zero.  You can't divide by zero, not even zero. 

$y = \frac{1}{3} x + \frac{5}{3}$

The answer is 1.

Apply the fraction rule: 4/3 x 3/4

Multiply the fractions: 4x3/3x4= 3/3

=1

Hope it helps. :)

Let the worth of the horse =H

[19200 + H] / 12 * 7 =10450 + H, solve for H

H =1,800 dollars - worth of the horse.

Note: You could also set up a system of equations as follows:

[19200 + H] =12M [M =Months]..........(1)

[10450 + H] =7M...................................(2), solve for H, M

H = 1,800 dollars - Horse's worth.

M =1,750 - his gross monthly compensation[19200/12 + 1800/12 =1,750]

How high is a tree that casts a 35​-ft shadow at the same time a 5​-ft pole casts a shadow which is 9​-ft​ long?   

                                               H            5  

This is a proportion:              –––   =   –––  

                                              35           9  

Do they still call it

cross-multiplying                     9H   =   (35)(5)  

                                                9H   =   175  

                                                  H   =   19.44  

_.

No, I just needed the answer to make sure I was correct. Thank you so much!

I got 8i as well so I think you're correct.  Do you need the steps?  

(x - 4) * (2y + 1)  = x * y + 1, solve for x, y

x = (8 y + 5)/(y + 1), substituting 1 for y does not give an integer.

But subbing 2 for y, we get:

x =(8*2 + 5) / (2 + 1) =21 / 3 =7

So, your numbers are: x = 7  and y = 2. And x * y =7 * 2 = 14

Your friend's numbers =(7 - 4) =3   and   (2*2 + 1) =5, And: 3 * 5 = 15

thats incorrect

Thanks!

140n  +  57, where n=0, 1, 2, 3......etc.

The smallest whole number = 57

This is also known as the "locker problem"

The number of  gerbils   who will  be wearing hats  at the  end will be number of perfect squares between 1 -104 inclusive =   10

To see this...let's suppose there are only 4  gerbils and  4  students

H = hat on     O  = hat off

                         Gerbils

Sudent 1        H H  H  H

Student 2       H O  H  O

Student 3       H O  O  O

Student 4       H  O O  H

Note  that after student 4  passes......the  number of gerbils wearing hats  are  just the  perfect squares from 1-4 inclusive =  2

This result can be generalized  to any N

Let x  be the  number  of $10 increases

Tickets sold * Price  = Revenue

We have this

(40000 - 800x) ( 200 + 10x)   = Revenue          simplify

40000(200) - (800 * 200)x + ( 40000*10)x + 8000x^2      = Revenue

8000000 - 160000x + 400000x  - 8000x^2  = Revenue

R  = -8000x^2  + 240000x + 8000000  

To maximize  the revenue....the number of $10   increases  = x =      -240000/(-2 * -8000)  =  15

So....the price of a ticket that maxes the revenue =  (200 + 15*10)  = 200 + 150  =  $350

(a + bi)  +  ( c + di)  + (e + fi)   = -2i

Subbing in we have

(a + 1i)  + ( c + di)  + ( -a - c + fi)  =  -2i

(a - a)  + (c - c) +  (1i + di + fi)

1i + di + fi  = -2i

di + fi =  -2i -1i

(d + f) i = ( -3) i

d + f =  -3

Here's my best effort

(1/2) the area  of  one of the  larger "leaves" is

The area of a  sector with a 120° arc  and a radius of  4  less the area of a triangle  with sides of 4 and an included angle of 120° =   

(1/2) 4^2 (2pi/3)  -  (1/2)(4)^2 sin (120°)  = 

(1/2) 16  ( 2pi/3 - √3/2)  =

8  ( 2pi/3 - √3/2)

And we have 6 of these areas  (2 per leaf)

So the area  of all three larger leaves =  6 * 8  ( 2pi/3  - √3/2)  = 48 (2pi/3  - √3/2)  = 96 pi/3 - 24√3   (1)

From this....we must subtract  the total area  of the  smaller  two "leaves"

(1/2)  of the area of  one of these  = 

Area of a sector with a 60° arc and a radius of 4 less the area of a triangle with sides of 4 and an included angle of 60° = 

(1/2)  (4^2) (pi/3)  - (1/2)(4^2) sin 60°  =

(1/2)(4^2)  [ pi/3  - √3/2] =

8 [ pi/3  - √3/2]

And we have 4 of these areas

So the total area of the smaller two leaves =   32 [ pi/3 - √3/2]  =  32pi/3 - 16√3     (2)

Total shaded area  =  (1) - (2)  =    96pi/3 -24√3  - [ 32pi/3 - 16√3  ]  =

[64pi/3  - 8√3]  units^2

19200  * (7/12)   = 11200  =  proportional pay for  7 months

Value of horse =   11200 -10450   = $750

x^2  - 10x + 36       complete the square on x....take 1/2 of 10  = 5.....square it.....=25   add and subtract this

x^2  - 10x + 25  + 36  - 25      factor the  first three terms 

(x - 5)^2  +  11

b = -5     c  =11

b + c   = 6

Equation of line

y = -(3/7)x  + 18

For x intercept, let y  = 0

0 = (-3/7)x + 18   subtract 18 from both sides

-18  = -(3/7)x    multiply both sides by - 7/3

(-18) ( -7/3)  = x  =   42  =  x intercept     

a)  The graph shows   the number of adults and youths that go to the movies on Friday, Saturday and Sunday

b)  The inference that  can be  made is that Friday  is a more  popular day  for  adults  while Saturday and Sunday are more popular days for youths

c)   

i)   Is true....Saturday is the  most popular day 

Attendance on Friday  = 650 + 300  = 950

Attendance on Saturday  =    (>500)  + 650   = > 1150

 Attendance on Sunday  =  (>300) + 500  = > 800

ii)   Adult attendance   650 + (>500) + (>300)  = >  1450

      Youth Attendance =  300 + 650  + 500 =  1450

Note that  (ii)  is also true.....the adult and youth attendance  are approximately equal

Here is the Aops way of doing it, same way as it did it:

Expanding the determinant, we obtain:

\begin{align*}
\begin{vmatrix} 1 & 4 & 9 \\ 3 & x & y \\ 3 & y & x \end{vmatrix} &= \begin{vmatrix} x & y \\ y & x \end{vmatrix} - 4 \begin{vmatrix} 3 & y \\ 3 & x \end{vmatrix} + 9 \begin{vmatrix} 3 & x \\ 3 & y \end{vmatrix} \\
&= (x^2 - y^2) - 4(3x - 3y) + 9(3y - 3x) \\
&= x^2 - y^2 - 39x + 39y \\
&= (x - y)(x + y) - 39(x - y) \\
&= (x - y)(x + y - 39).
\end{align*}

Since this is 0, either x - y = 0 or x + y - 39 = 0.  But x and y are distinct, so \(x + y = \boxed{39}.\)

First series......common ratio between terms is  4/12 = 1/3

Sum of this series is  represented by :       12 / [ 1 -1/3 ] =  12/ (2/3)   = 18

The second series has a sum of  18 * 3  = 54

The common ratio is  [4 + n] /12

So

12 / [ 1 - (4 +n)/12 ]  =  54

12/54  =  1  - (4 + n)  /12

2/9 = 1  - (4 + n) /12

-7/9  = - (4 + n) /12

12 (7/9) = 4 + n

84/9 - 4   = n

48/9 = n  =  16/3  [just as the Guest found!!!  ]

Solve 3sin(x-y)sin(x+y)=0

Hello Guest!

\(3sin(x-y)sin(x+y)=0\\ 3(sin(x)cos(y)-cos(x)sin(y))(sin(x)cos(y)+cos(x)sin(y))=0 \)

               u                           v                         u                           v

\(3(u-v)(u+v)=0\\ 3(u^2-v^2)=0\\ u^2=v^2\\ u=v\)

\(\)\(y=90°-x\)

  !

Notice  that   4sqrt (1/2)  =    ±  4 (1) / sqrt (2)  =  4/sqrt (2)

Multiply  top/bottom  by  sqrt (2)   and we get

 ±  4sqrt (2) / 2 =   ±  2 sqrt (2)  =  b

Sorry that I didn't simplify it  more  !!!