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\( x^2+y^2=2x-6y+6+14x-14y+28\)

This is just a circle. i moved the xs and ys to the left and that just moves the circle. the size depends on the constant which is 34. so the area of the circle would be 34pi

Thanks! 

LCM  =  2^4 * 3^2 * 5 * 7 * 11* 13

2 = 2                   9  = 3^2               16 = 2^4

3 = 3                   10 = 2 * 5

4 = 2^2               11 = 11

5 = 5                    12 = 2^2 * 3

6 = 2 * 3               13 = 13

7 = 7                     14 = 7 * 2

8 = 2^3                 15 = 3 * 5

LCM = 2^4 * 3^2 * 5 * 7 * 11 * 13

# positive divisors =  (4 +1) (2 + 1) (1 + 1)^4  =  240 

When we are trying to find the roots, we set q(y) to equal 0. 

Thus, we have

\(12y^3-28y^2-9y+10=0\)

Factoring the polynomial, we have

\(12y^3-28y^2-9y+10=0\)

Thus, we have

\(y=\frac{1}{2},\:y=-\frac{2}{3},\:y=\frac{5}{2}\)

Thanks! :)

Try to use polynomial division by dividing values of (x +or- \(\frac{divisors\space of \space 12}{divisors\space of \space 10}\))

im wayyyyy too lazy to do it. but eventually by trying all values you will get a quadratic. then it becomes mucchhhh easier :)

\(x^2 + nx + 15 < -89 - 3x^2\)

so i first brought everything to the left side to get \(4x^2 + nx + 104 < 0\)

then i divided by 4 to get\(x^2 + \frac{n}{4}x + 26 < 0\)

for x to have no real solutions the discriminant must be negative

so \(\frac{n^2}{16}-4(26)\) must be negative. the less than sign is useless since smaller right hand values make c greater

the greatest value of n that works is 40

but we also can have negatives, least of which is -40

so -40 to 40 is \(\boxed{81}\) numbers

I know there are smarter ways to do it...but I'm not smart...as my name suggests

We could simply just take a look at the factors. 

\((2u - 3*v^3)*5*(3u^3 - 2v^5)^3\) never has a term for v^9, so it's impossible. 

We can easily tell, but ummm...yeah....

XD :)

~NTS

First, let's move all terms to one side and simplify. We have

\(4x^2 + nx + 104 < 0 \)

Let's set this up as an equation. We have

\(4x^2 + nx + 104 = 0\)

This will have real solutions when

\(n^2 - 4 (4) (104) ≥ 0 \\ n^2 ≥ 1664\)

\(n ≥ 40.79 \\ n ≤ -40.79\)

This is when there is a real value. Which means that when 

\( n < 40.79 \\ n > -40.79\)

there will be no real solutions. Thus, we have 

\( n = [ -40 , 40 ] \)

There are 81 values of n. 

Thus our answer is 81 !

Thanks! :)

Depends on the factorization of m

m =  2 * 3^8   has  (1 + 1) (8 + 1)  = 18 divisors

m^2  = [ 2 * 3^8]^2  =  2^2 * 3^16  has ( 2 + 1) (16 + 1)  =  51 divisors

But

m = 2^2 * 3^5  has (2+1)(5 + 1)  = 18 divisors

m^2  = [ 2^2 * 3^5]^2  = 2^4 * 3^10  has (4 + 1) (10 + 1)  =  55 divisors

And as  NTS pointed out, 

m = 2^2 * 3^2 * 5  has (2 + 1) (2 + 1) (1 + 1)  =18 divisors

But

(2^2 * 3^2 * 5)^2 = 2^4 * 3^4 * 5^2   has ( 4 + 1 )(4 + 1) (2 + 1)  = 75 divisors

\((x^2 + 1/x)*(x + x^4)^3\)

i then reformatted stuff to get\((\frac{x^3+1}{x})(x^3+3x^6+3x^9+x^{12})\)

i further expanded to get\(\frac{x^3+4x^9+6x^9+4x^{12}+x^{15}}{x}\)

then just substract one from all the powers in the numerator to get the final answer of \(\boxed{x^2+4x^5+6x^8+4x^{11}+x^{14}}\). :)

Let's put them all of them with denominator x. 

\(\frac{\left(x^3+1\right)\left(x+x^4\right)^3}{x}\)

Factoring the cube first, we get

\(\frac{\left(x^3+1\right)x^3\left(x+1\right)^3\left(x^2-x+1\right)^3}{x}\)

Canceling out the denominator, we get

\(\left(x^3+1\right)x^2\left(x+1\right)^3\left(x^2-x+1\right)^3\)

Expanding everything, we finally get

\(x^{14}+4x^{11}+6x^8+4x^5+x^2\)

That's our final answer!

Thanks! :)

XD just expanding is definentaly(100% misspelled) a way to do it

The smallest possible positive integer value of x would be 1, since 5 + 2*-2 =1. The answer to part one of your question is \(\boxed{1,-2}\)

im pretty sure the euclidean algorithm gives infinite solutions. after i did the process x=1+2k, y=−2−5k. so you just plug in values of k to get solutions. Im not sure what the question is asking but im pretty sure it wants \(\boxed{x=1+2k,\space y=−2−5k}\). 

I basically expanded everything out. 

\(120v^{18}-540u^{3}v^{13}-80uv^{15}+810u^{6}v^{8}+360u^{4}v^{10}-405u^{9}v^{3}-540u^{7}v^{5}+270u^{10}\)

hmmm...there is no u^2v^9 term. 

So the coefficient is just 0. 

So 0 is our answer. 

Thanks! :)

well for an integer to have 8 positive divisors, it means that its prime factorization is either p*q^3 or p*q*r, where p,q, and r are primes. uhhh its literally imposible to determine all the positive integers with this prime factorization since there are infinite numbers of primes. :P

You have a total supply of $1000$ pieces of candy, and an empty vat.  You also have a machine that can add exactly $7$ pieces of candy per scoop to the vat, and another machine that can remove exactly $6$ pieces of candy with a different scoop from the vat.  When these two machines are done, there is only one piece of candy left in the vat.  What is the smallest possible number of times the the first machine added candy to the vat?  

Well, I say the least possible number number of times is 1 time.  

Turn on both machines.  Machine 1 adds 7, and Machine 2 removes 6,   

then real quick turn off both machines before they have time to cycle again.  

Nobody said the machines have to keep running until all 1000 pieces are gone.    

_.

_.

_.

I'm not sure how to explain the prcoess, but the smallest number with 18 positive divisors is \(180\)

We have \(1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180.\)

Now, \(180^2 = 32400\)

You could also use prime factorization, by doing

\(32400 =2 ^ 4 ×3 ^ 4 ×5 ^ 2\)

\((4+1)(4+1)(2+1)=5×5×3=75\)

Thus, our answeris 75. 

Thanks! :)

Let's note that 1/17 is a repeating decimal, which can make our lives quite easy!

We have

\(1/17 = 0.\overline{0588235294117647}\)

There are 16 repeating digits. Thus, we simplfy have

\(4000/16=250\)

Thus, the last digit, 7 is the 4000th digit. 

So our answer is 7. 

Thanks! :)

First, let's express a^3 in terms of ab^2. We have

\( a^3 = 679 - 3ab^2 \)

Now, let's subsitute this value for a into the second equation. We get that

\( 3(679 - 3ab^2) - ab^2 = 615 \)

Now, we simply solve for ab^2. We have that

\(\\ 2037 - 9ab^2 - ab^2 = 615 \\ \\9ab^2 + ab^2 = 2037 - 615 \\ 10ab^2 = 1422 \\ ab^2 = (1422)/(10) \\ ab^2 = 142.2 \\\)

Subsituting this value back into the first equation, we have

\( a^3 + 3 * 142.2 = 679 \\ a^3 + 426.6 = 679 \\ a^3 = 679 - 426.6 \\ a^3 = 252.4 \\ a = \sqrt[3]{252.4} \\ a \approx 6.19 \\\)

Now, we solve for b. We get

\(b^2 = (142.2)/(6.19) \\ b^2 \approx 22.97 \\ b \approx \sqrt{(22.97)} \\ b \approx 4.79 \\\)

Thus, 

\( a - b \approx 1.40 \)

So our answer is approximately 1.4

Thanks! :)

First, let's convert the numbers in to base 19. 

We have

\((F3)_{19} = (15 × 19^1) + (3 × 19^0) = (288)_{10 }\)

This means that \(\frac{1}{288}_{10}=\frac{1}{F3}_{19}\)

Doing some base division, we have

\(\frac{1}{F3}_{19}=\overline{0.014F9AE6}\)

Thus, it is a repeating decimal. 

So our answer is repeating. 

Thanks! :)

My pleasure! 

I hope you understood! :)

Thanks! :)

I want a helpful solution, I'm not trying to brute force it with Wolfram Alpha. Thanks though!

By the integer root theorem, the an integer root must be a factor of 72, like 1, 2, 3, 4, 6, ...  Checking these, we find that 18 is an integer root.