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There are two possible cases:

Case 1: The downtown location has a square display. In this case, the number of pies in the downtown location is 44 = 16. The number of pies in the uptown location is 66 = 36. Therefore, there are 6 pies in each row of the uptown location.

Case 2: The uptown location has a square display. In this case, the number of pies in the uptown location is 66 = 36. The number of pies in the downtown location is 44 = 16. Therefore, there are 4 pies in each row of the downtown location.

We cannot know which location has the square display, so both cases are possible. Therefore, the answer is either 6 or 4.

Good answer.

Hi guest,

An aquarium tank is 1/6 full of water. When 2 gallons of water are added, the tank becomes 1/2 full. What is the total capacity of the aquarium tank, in gallons?  

Note that 1/2 is the same as 3/6.  

So 2 gallons took the tank from 1/6 to 3/6. 

So 2 gallons added 2/6.  That means 1 gallon is 1/6.  

Therefore, the total capacity of the aquarium is 6 gallons.  

_.

In Linguistics 101, the ratio of the number of juniors to the number of seniors is 3:2.  When six more juniors join the class, and one senior drops the class, the ratio of the number of juniors to the number of seniors becomes 2:3.  How many students are in the class after these changes?  

                                                        J          3  

                                                       ––   =   ––  

                                                        S         2  

                                                       2J  =  3S  

                                                      (J + 6)         2  

                                                      –––––   =   ––   

                                                      (S – 1)        3  

                                                      3J + 18  =  2S – 2  

                                                        J  =  3S/2  

                                                       (3)(3S/2) + 18  =  2S – 2  

                                                        9S/2 + 18  =  2S – 2        

Multiply everything by 2 to    

get rid of the denominator                   9S + 36  =  4S – 4  

 I saw here that the number of Seniors is going to be negative  

I went back and considered the original proposal.    

If you've got Juniors to Seniors = 3 to 2  then add more Juniors  

than Seniors, the proportion could change, but it can not invert. 

This problem cannot be solved.  

_.

This might help:

Let's solve the system of equations step by step to find the ordered triple (p, q, r) that satisfies the given equations:

Equation 1: p - 2q = 3

Equation 2: q - 2r = -2 + q
Simplifying: -2r = -2

Equation 3: p + r = 9 + p
Simplifying: r = 9

From Equation 2, we found that -2r = -2. Dividing both sides by -2 gives us r = 1.

Now, we can substitute the values of r into the other equations to find the remaining variables:

From Equation 3: p + 1 = 9 + p
Simplifying, we see that 1 = 9, which is not true. Therefore, there is no solution for p in this system of equations.

Since there is no solution for p, q and r cannot be determined. The system of equations is inconsistent and does not have an ordered triple (p, q, r) that satisfies all the equations simultaneously.

Hi HumenBeing,

We have two equations in two variables.

So, from the first equation: \(3x+4y=15\), we can choose to make whichever variable the "subject" (I.e. Isolated)
So, for example, I choose x:  \(x=\dfrac{15-4y}{3}\)

Next, we substitute this in the second equation: \(x^2+y^2=9\)

Therefore, \((\dfrac{15-4y}{3})^2+y^2=9 \\ \iff \dfrac{1}{9}(15-4y)^2+y^2 = 9 \space \space \space \space \text{Focus more on this step} \\ \iff \dfrac{1}{9}(225-120y+16y^2)+y^2=9 \\ \text{It would be much easier if we got rid of fractions, so why don't we multiply both sides by 9?} \\ \iff 225-120y+16y^2+9y^2=81 \\ \iff (225-81)-120y+(16+9)y^2=0 \\ \iff 144-120y+25y^2=0 \\ \iff 25y^2-120y+144=0 \\ \text{This is a Quadratic Equation, we can solve it either by factoring or by the quadratic formula} \\ \text{But, notice this is a perfect square:} \\ (5y-12)^2=0 \\ \iff y=\dfrac{12}{5} \text{Now, we have found the only value of y} \\ \text{To find the x-coordinate, we just substitute this value of y in the very first equation where we isolated "x"} \\ \implies x=\dfrac{15-4y}{3}=\dfrac{15-4*\dfrac{12}{5}}{3}=\dfrac{9}{5} \)

Therefore, the point is: \((\dfrac{9}{5},\dfrac{12}{5})\)

To check this answer, go to desmos.com and write both equations, notice the line is tangent to the circle at (1.8,2.4) which is exactly the point that we have just found!  

I hope this helps.

I'm sorry... that's not correct... but thanks for trying!

The graph of the equation 3x + 4y = 15 is a line. The graph of the equation x^2 + y^2 = 9 is a circle. The lines and circle intersect at two points.

To find the coordinates of these points, we can solve the system of equations.

We can do this by first subtracting 3x from both sides of the equation 3x + 4y = 15. This gives us:

4y = 15 - 3x

We can then divide both sides of this equation by 4. This gives us:

y = \frac{15}{4} - \frac{3}{4}x

We can then substitute this equation for y in the equation x^2 + y^2 = 9. This gives us:

x^2 + \left( \frac{15}{4} - \frac{3}{4}x \right)^2 = 9

We can then simplify this equation. This gives us:

x^2 + \frac{225}{16} - \frac{45}{2}x + \frac{9}{16} = 9

We can then combine constant terms on the left-hand side of the equation. This gives us:

x^2 - \frac{45}{2}x + \frac{162}{16} = 0

We can then factor the left-hand side of the equation. This gives us:

\left( x - \frac{27}{4} \right)^2 = 0

We can then take the square root of both sides of the equation. This gives us:

x - \frac{27}{4} = 0

We can then solve for x. This gives us:

x = \frac{27}{4}

We can then substitute this value of x into the equation y = \frac{15}{4} - \frac{3}{4}x to find the y-coordinate of the point of intersection. This gives us:

y = \frac{15}{4} - \frac{3}{4} \cdot \frac{27}{4}

This simplifies to:

y = -\frac{9}{2}

Therefore, the point of intersection of the two graphs is (27/4, -9/2).

Thank you all!

2020 ==F / [1 - r],       

20200=F^2/[1 - r^2], solve for F, r

Divide the 2nd by the first:

F ==4040 / 203 - this is the first term of the original sequence

r ==201 / 203 - this is the common ratio of the original sequence.

None of your LaTex is displaying properly for me.

No, it isn't. n=6. Expand: (4x^2 - 7y^3)^6 ==4096 x^12 - 43008 x^10 y^3 + 188160 x^8 y^6 - 439040 x^6 y^9 + 576240 x^4 y^12 - 403368 x^2 y^15 + 117649 y^18 - where is:x^2 y^6

n==3.  Expand:  (4x^2 - 7y^3)^3 ==64 x^6 - 336 x^4 y^3 + 588 x^2 y^6 - 343 y^9

The term cx^2y^6 will be present in the expansion of (4x^2 - 7y^3)^n if and only if n is divisible by 2 and 3. This is because the term cx^2y^6 can only be created by combining two factors of x^2 and three factors of y^3.

The smallest positive integer n that satisfies these conditions is n = 6. This is because 6 is divisible by 2 and 3, and the expansion of (4x^2 - 7y^3)^6 contains the term 4x^2y^6.

Here is a more detailed explanation of why n must be divisible by 2 and 3.

The term cx^2*y^6 can only be created by combining two factors of x^2 and three factors of y^3. This means that there must be at least two instances of x^2 and at least three instances of y^3 in the expansion of (4x^2 - 7y^3)^n.

The expansion of (4x^2 - 7y^3)^n contains a term of x^2 in the first term, which is 4x^2. This term will be present in the expansion regardless of the value of n.

The expansion of (4x^2 - 7y^3)^n contains a term of y^3 in the third term, which is -7y^3. This term will be present in the expansion regardless of the value of n.

For the term cx^2*y^6 to be present in the expansion, there must be at least one other term in the expansion that contains a factor of x^2 and a factor of y^3.

The only way for this to happen is if n is divisible by 2 and 3. If n is not divisible by 2, then there will be no other terms in the expansion that contain a factor of x^2. If n is not divisible by 3, then there will be no other terms in the expansion that contain a factor of y^3.

Therefore, n must be divisible by 2 and 3 in order for the term cx^2*y^6 to be present in the expansion.

The question isn't the same tho in that link. 

Good answer.

Anyways,      \(\boxed{n=3}\)

\(x=120\)                                            \(240000=2000x \)                   \(\dfrac{2000}{60}=\dfrac{6000}{x}\)