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x + y = 10         square both sides

x^2 + 2xy + y^2 = 100

x^2 + y^2  = 100 - 2xy

So

100 - 2xy  = 64 + xy

100 - 64 = 3xy

36 = 3xy

12 = xy →   y = 12/x

So

x + 12/x = 10

x^2 -10x + 12  =  0

x = [ 10 +/- sqrt [100 - 48] ] / [ 2  =  [ 10 +/- sqrt [ 52] ]/ 2  =  [10 +/- 2sqrt13 ] / 2  = 5 +/- sqrt 13

Using the conjugate property

(x,y) =  ( 5 + sqrt 13 , 5 -sqrt 13 )  ,  (5 -sqrt 13 , 5 + sqrt 13)

mmm

\(\frac{x}{x - 5} = \frac{4}{2x - 4} + \frac{2}{x + 3}\)

for starters x cannot equal  5, 2 or  -3

\(\frac{x}{x - 5} = \frac{4}{2x - 4} + \frac{2}{x + 3}\\ \frac{x}{x - 5} = \frac{2}{x - 2} + \frac{2}{x + 3}\\ x(x-2)(x+3)=2(x-5)(x+3)+2(x-5)(x-2)\)

Now expand out and see where it goes.

It did not fall out for me.

Thanks Pythagorean.

I think I will try this one too.

I'll use a method called stars and bars.

Put the 12 stickers in a row (they are the stars)

Now we need to divide them into 4 groups.  We can do this using 3 bars. 

eg  X X | X X X X | | X X X X X X        represents the outcome of   2, 4, 0, 6

The question now becomes:  How many positions can the 3 bars be placed in

The answer is    15C3 = 455 ways

Our answers are close Pythagoras but I think you have double counted some. ..

I got 36,  it might not be right.

There are only 3 ways to pair the siblings so that no two siblings are together.

But then either person can be in front of the pair so Ill double it to 6

The first pair has a choice of 3 seats in the row, the second pair has a choice of 2 and the last has the choice of 1

So that is    3*2*3*2*1 = 36

x + y =  6     square both sides

x^2 + 2xy + y^2 = 36

x^2 + y^2  = 36 -2xy

x^3 + y^3  = 162 - (x^2 + y^2)

(x + y) (x^2 + y^2 - xy) =  162 - (x^2 + y^2)

   6 (x^2 + y^2 - xy) = 162 -(x^2 + y^2)

 6 (36  - 2xy -xy) = 162 - (36 -2xy)

216 - 18xy = 126 + 2xy

90 = 20xy

xy = 9/2

y = 9/(2x)

x + y = 6

x + 9/(2x) = 6

2x^2 + 9 = 12x

2x^2 -12x + 9 = 0

x =  [12 + sqrt [ 144 - 72] ] / 4  =  [ 12 + 6sqrt2] / 4  =  3 + 3/sqrt 2

y =  3 - 3/sqrt 2

When we have f(x+3), the graph shifts to the left by 3 units. 

When we have -8, the graph shifts down 8 units. 

The vertex will not be affect by the 3 and the 2. 

Thus, the vertex of the new parabola is (-2, -10) when we shift it 3 units to the left and 8 units down!

Thanks! :)

First, let's combine some like terms!

\(x^2 - 22c + c\). 

In order for this to be the square of a binomial, we need c to be the square of half the value of -22. 

Solving this, we get c = 121. 

Indeed, if we have \(x^2-22x+121=(x-11)^2\). 

So we have c = 121. 

Thanks! :)

First off, let's combine like terms to get \(x^2 + sx + 100\). 

For this to be a square of a binomial, we must recognize that s is equal to double the square root of 100. 

Solving this we get s=20. 

Indeed, \(x^2+20x+100 = (x+10)^2\). 

However, we also get s=-20. 

We have \(x^2-20x+100 = (x-10)^2\). 

This means s=-20, 20

Thanks! :)

(Haha, this is the post that got me to 200 points!)

Alright! I'll tackle this problem!

First off, we notice there is an f(x) function inside the second equation. 

We know f(3)=4, we let's let x/3 in the second equation be 3 so that we have an integer. 

Setting x=9 would get us this desired equation. 

We have \(y=2f(3) + 17\), which is essentially \(2(4)+17\). 

This simplifies to 25, meaning when x equals 9, we have y equals 25. 

This means that (9, 25) HAS to be on the graph for y=2f(x/3)+17!

Thanks! :)

We can use constructive counting or casework to solve this problem. 

CASE 1

First, let's see all the two digit numbers that satisfy these conditions. 

We only have two, with 23 and 32. 

CASE 2

Now, let's check all the 3 digit numbers. 

We have 311, 131, 113, giving us 3. 

We also have 221, 212, and 122, giving us another 3. 

Case 3

4 digit number time! This is where it gets a bit complicated, but it's still not that bad. 

We have numbers with three 1s and one 2, which has 4 different possibilities. 

Case 4

Last case ! 5 digit numbers only have 1, with 11111. 

Now, we just have to add up all our cases, and we get 2+3+3+4+1 = 13. 

We have 13 numbers that satisfy the conditions. 

NOTICE: This problem didn't have a lot of cases, so it would be easy to count. Bigger and wider ranges would require permutations, which I really really dislike! :)

I hope I answered your question!

Thanks! :)

This might seem like a weird way to do it, but like others, I would reccommend the Cauchy-Schwarz inequality.

Essentially, what this inequality states is that where a and b are real numbers, we can write the inequality \((a_1^2+a_2^2+\dots+a_n^2)(b_1^2+b_2^2+\dots+b_n^2)\ge(a_1b_1+a_1b_2+\dots+a_nb_n)^2\)

We could apply this to this problem!

We have \((x^2+3y^2)(1+\frac{1}{3}) \geq (x+\sqrt{3}(\frac{1}{\sqrt{3}})y)^2\). 

\(18*\frac{4}{3}\ge(x+y)^2\)

\(x+y\le\sqrt{24}\)

Clearly, from this equation, we find that the maximum value is \(\sqrt{24}\)

We could probably brute force this problem, but this trick is defintely easier!

Thanks! :)

First, let's simplify the expression by combining like terms. 

\(x^2 -14x + y^2 + 16y + 26\). 

Ok! Now all we have to do is to take the derivative with respect to x and y and set that to 0!

We have 

\(2x - 14 = 0 \\ 2x = 14 \\ x = 7\)

and 

\(2y + 16 = 0 \\ 2y = -16 \\ y = -8\)

Now, all we have to do is to plug in these values for x and y into the orginial function. 

We get \((7)^2 - 14(7) + (-8)^2 + 16(-8) + 26 = -87\). 

The lowest temperature on the plane is -87 degrees, which is quite cold!

Thanks! :)

First, let's combine some like terms to simplify this expression!

\(x^2 -14x + y^2 + 16y + 26 \)

Now, all we have to do is to take the derivative with respect to x and y and set that to 0 before plugging the values back into the function. 

We have 

\(2x - 14 = 0 \\ 2x = 14 \\ x = 7\)

and 

\(2y + 16 = 0 \\ 2y = -16 \\ y = -8\)

Now, plugging the value back into the function, we have \((7)^2 - 14(7) + (-8)^2 + 16(-8) + 26 = -87 \). 

So the lowest temperature is just -87!

Thanks! :)

First, let's isolate y in the first equation. 

\(xy = 2/5 \\ y = 2 / (5x) \)

Plugging this value into the second expression, we get 

\(6x + 3 / (2 / (5x) ) = \\ 6x + (15/2)x = \\ (27/2) x\)

However, notice that there is no mimumum value or maximum value because the smaller x is, the smaller the number is to no limit. 

There are absolutely no restrictions for this expression. 

Thanks! :)

Let's use casework!

We can consider different cases based on how many friends receive stickers:

No friends receive stickers: There is only 1 way to do this (give none to anyone).

One friend receives stickers: There are 4 ways to choose which friend gets stickers (out of the 4 friends), and then there is 1 way to distribute all 12 stickers to that one friend (since they are identical). So, there are 4 * 1 = 4 ways for this case.

Two friends receive stickers: There are 4 ways to choose which two friends get stickers, then we can distribute the stickers in (x12​) ways, where x is the number of stickers given to the first friend (the remaining 12 - x stickers go to the second friend).

The number of ways to distribute the stickers like this is the same for any choice of which friend gets x stickers, so we sum over all possible values of x from 1 to 11. This gives us a total of:

4 * (11 + 10 + ... + 1) = 4 * \frac{11 \cdot 12}{2} = 264

Three friends receive stickers: Similar to the case of two friends, there are 4 ways to choose which three friends get stickers, and then we can distribute the stickers in (x12​) ways for various values of x (number of stickers to the first friend). Summing over all possibilities for x gives us:

4 * (11 + 10 + ... + 2) = 4 * \frac{11 \cdot 10}{2} = 220

Four friends receive stickers: There are 4 ways to choose which four friends get stickers, and then 1 way to distribute all 12 identical stickers among them. So, there are 4 * 1 = 4 ways for this case.

Adding the number of ways for each case, we get a total of:

1 + 4 + 264 + 220 + 4 = 493

Therefore, there are 493 ways for Magnus to distribute his stickers.

Since angle EDF is inscribed in a circle, we know that the measure of the central angle subtended by arc DF is double the measure of angle EDF. Therefore, the central angle for arc DF is 29 x 2 = 58 degrees.

Since the circumference of the circle is 24 units and the central angle for arc DF is 58 degrees, we can use the formula for the circumference of a circle to find the radius of circle G.

C = 2πr
24 = 2πr
r = 12/π

Now, we can use the formula for the length of an arc, given by:

Length of arc = (central angle/360) * circumference

For arc DE, the central angle is 360 - 132 - 58 = 170 degrees.

Length of arc DE = (170/360) * 2π(12/π)

Length of arc DE = (17/36) * 24

Length of arc DE = 8/3 units

Therefore, the length of arc DE is 8/3 units.

We are given that the circumference of circle G is 24 units, so the radius of circle G is 24/2π=12/π units.

An inscribed angle intercepts an arc that has a measure half that of the central angle that subtends the same arc [Inscribed Angle Theorem]. Therefore, central angle ∠EDF measures 2⋅29∘=58∘.

The measure of arc DF is given as 132 degrees. Since the sum of the degrees around a circle is 360∘, the measure of arc DE is 360∘−58∘−132∘=170∘.

The ratio between the arc's central angle θ and 360∘ is equal to the ratio between the arc length s and the circle's circumference c [Arc Length Ratio Theorem].

In this case, the ratio of the arc DE's central angle θ (which is 170∘ ) to 360∘ is equal to the ratio of the arc DE's length s (what we want to find) to the circle's circumference c (which is 24 units).

Therefore:

360∘θ​=cs​

360∘170∘​=24 unitss​

Solving for s, we get the arc length s of arc DE:

s=360∘170∘​⋅24 units = 17/3 units

Therefore, the answer is 17/3.

Laverne starts counting out loud by 5's. She starts with 1. As Laverne counts, Shirley sums the numbers Laverne says. When the sum finally exceeds 40, Shirley runs screaming from the room. What number does Laverne say that sends Shirley screaming and running? 

Laverne's           Shirley's  

Counting          Summation  

     1                           1  

     6                           7  

   11                         18   

.  16                         34  

   21                         55    

_.

Chuck deposits $2000 into a bank account that compounds weekly at an annual interest rate of 15% Assuming there are no other transactions, what will the balance be after 10 years, in dollars?  

Since the yearly interest rate for Chuck's account is 15% (same as 0.15) and there  

are 52 weeks in a year, then the amount earned the first week is (2000)(0.15 / 52)  

I'm going to change (0.15 / 52) to its decimal equal.  I'll round that value to 0.00288   

only for the convenience in writing it in this post.  In my calculator I will not round it,  

but maintain the entire 15-digit decimal in the calculations.  

Since the bank is adding (0.00288) of the current balance to the account each week,  

the total in the account (principal plus interest) after the first week is (2000)(1.00288).  

Then each week, the balance is (1.00288) times what it was the week before.  

Let's make a little tabulation, to demonstrate clearly the progression of the balance.   

Balance at end of 1^st week      (2000)(1.00288)  

Balance at end of 2^nd week     (2000)(1.00288)(1.00288)  

Balance at end of 3^rd week      (2000)(1.00288)(1.00288)(1.00288)   

Balance at end of 4^th week      (2000)(1.00288)(1.00288)(1.00288)(1.00288)   

Balance at end of n^th week      (2000)(1.00288)ⁿ   

Balance at end of 520^th week   (2000)(1.00288)⁵²⁰ =  8944.04 dollars   

_.

Problem 3:

To find the integers n that satisfy 64≤n2≤100, we can take the square root of both sides of the inequality. However, since the square root function is not one-to-one (it has two outputs for each positive input), we need to be careful.

Taking the square root of both sides gives:

8≤∣n∣≤10

Since we only care about positive integer solutions for n, we can simply remove the absolute value signs and consider 8≤n≤10.

Now, we can count the number of integers in this range, inclusive. There are 10−8+1=3​ integers that satisfy the inequality: 8, 9, and 10.

Problem 2:

To find the largest possible value of a + 2b + 3c, we want to maximize the values of a, b, and c.

Since a, b, and c can be the numbers 1, 2, 3 in some order, let's consider the case where a = 3, b = 2, and c = 1:

a + 2b + 3c = 3 + 2(2) + 3(1) = 3 + 4 + 3 = 10

Therefore, the largest possible value of a + 2b + 3c is 10.

Problem 1:

We can complete the square to minimize the given expression

Steps to solve:1. Complete the square

First, we recognize that the expression is already in the form of a perfect square trinomial up to a constant factor. This means we can complete the square by taking half of the coefficient of our x2 term, squaring it, and adding it to both sides of the equation.

In this case, the coefficient of our x2 term is −2, so half of it would be −1, and squaring it gives us 1. Because we're adding the constant term inside the parentheses on the right where it's being multiplied by 1 (the coefficient of x2 ), we need to add −1 (the opposite of half the coefficient of our x2 term squared) outside the parentheses on the left to make sure we're adding the same thing to both sides.

x4−2x2−1=x4−2x2−1

(x2)2−2(x2)+1=(x2−1)2

2. Minimum value:

Now that the expression is a squared term, we know its minimum value is 0, which occurs when x2=1 or x=±1.

Answer: The minimum value of the expression is x2(x2−2) is 0,​ which occurs when x=±1.

For problem 3, the answer is 3/2.