NotThatSmart

Last question of the day for me. 

First, let's simplify the equation. Bringing all terms to on side and combining all like terms, we get

\(x^2 - mx + 14 = 0 \)

Now, we simply search for factors of 14, both negative and positive. We have

Factors of 14                                                m

(1 , 14)                   (x + 1) (x + 14)              -14

(-1, -14)                  (x - 1)  (x - 14)               14

(2,7)                       (x + 2) ( x + 7)                -9

(-2, -7)                   (x - 2) ( x - 7)                   9

4 should be the answer. 

Thanks! :)

Combining like terms and moving all terms to one side, we get

\(x^{2}+12x-4=0\)

Using the quadratic equation, we get

\(x=\frac{-12\pm \sqrt{12^{2}-4\cdot 1(-4)}}{2\cdot 1}\)

Finally simplifying, we get

\(x=2\sqrt{10}-6\\ x=-2\sqrt{10}-6\)

Thanks! :)

First, let's simplify the equation so that all terms are on one side.

Combining all like terms and moving them all to one side, we get the equation

\(x^2 + (17 - m)x + 4 = 0 \)

Now, since it has two distinct roots, the descriminant of the quadratic must be greater than 0.

Thus, we have the equation

\((17 - m)^2 - 4(1)(4) > 0 \\ (17 - m)^2 > 16 \)

We get two roots from this equation. 

Let's calculate both of them. For the first one, we have

\(17 - m > 4 \\ 17 - 4 > m \\ m < 13 \)

For the second root, we have

\( 17 - m < -4 \\ 21 < m \\ m > 21\)

Thus, in interval notation, we have \(m = ( -\infty, 13) U ( 21, \infty) \)

Thanks! :)

Ok, so we start off with the equation

\(ab -3a + 4b = 131 \)

I know this is random, but let'st ake the product of the coefficients on a,b.....add this product to  both sides

So this gets us

\(ab - 3a + 4b + (4 * - 3) = 131 + (4 * -3) \\ ab - 3a + 4b - 12 = 119 \)

This eventually achieves the equation

\((a + 4) ( b - 3) = 119\)

Now, the factors of 119 are \(1,   7 ,   17    , 119\)

Since we want the minimum, let's use 7 and 17. We have

\(( 13 + 4) ( 10 - 3) → | a - b | = |13 - 10 | = 3\)

Thus, our final answer is 3. 

Thanks! :)

Let's create a system of equations to solve this question. 

First, we know that the two points must be on the equation \(x^2 + y^2 = 25\)

We also know that x must be 4. 

Thus, plugging in x=4 into the first equation, we can find y values. We have

\(4^2+y^2=25\\ y^2=25-16\\ y=\pm\sqrt9\\ y=3, y=-3\)

Thus, the two points A and B are \((4,3), (4,-3)\)

Since they share an x value, the difference between the y values is the distance. We have

\(3-(-3)=6\)

Thus, 6 is our final answer. 

Thanks! :)

We can use Euler's Totient Function. 

Let's find all the distinct prime factors of 2835 first. 

The prime factors of 2835 are 3, 5, and 7

Now, we simplfy do \(2835(1-1/3)(1-1/5)(1-1/7)\)

Simplfying this, we have

\(2835 * \frac{2}{3}*\frac{4}{5}*\frac{6}{7} =\frac{2835\cdot \:2\cdot \:4\cdot \:6}{1\cdot \:3\cdot \:5\cdot \:7}=1296\)

So 1296 is our answer. 

I'm not sure if I did this correctly...

Thanks! :)

Let's use a whole number of equations to solve this problem. 

First off, we know that \((a+b)^2=a^2+2ab+b^2\)

We ALSO know that \((a+b)^2=4^2=16\)

Thus, we have the equation \(a^2+2ab+b^2=16\)

We are given a^2+b^2 in the problem, so plugging that in, we get

\(6+2ab=16\)

We can now find ab. This will come in handy later. 

\(2ab=10\\ ab=5\)

Alright, let's move on to what we are TRYING to find. 

Let's note that we can split a^3+b^3 into 

\(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

WAIT! We already know the values of every number! Plugging in 4, 6, and 5, we get

We have

\((4)(6-5)=4\)

So 4 SHOULD be the answer. 

I might have a mistake during the calculations...not sure. 

Thanks! :)

(b) Let's use a different tactic here. 

We want the fact that 4x subtracted by the remainder is divisble by 75. 

We can have the equation \(75y=4x-71\), where y is another integer. 

Isolating x from this equation, we get

\(x=\frac{75y+71}{4}\)

Now, any value of y we plug in should work. We just need an integer, and we're done. 

When we plug in \(y=3\), we find that

\(x=\frac{75(3)+71}{4}=\frac{296}{4}=74\)

Thus, 74 is our answer. Just fit it into the interval! Lol

Thanks! :)

(a) Let's note a pattern in the equation. 

\(x=1; 346 \equiv 346\pmod{347}\\ x=2; 346*2 \equiv 345\pmod{347}\\ x=3; 346*3 \equiv 344 \pmod{347}\\ x=4; 346*4 \equiv 343 \pmod{347}\\ etc. \)

Thus, we have 

\(347-x=129\) as the pattern for this congruence. 

thus, solving for x, we find that \(x=218\)

Testing if this is indeed true, we find that 

\(x=218; 346*218 \equiv 129 \pmod{347}\)

So 218 is the answer. 

Thanks! :)

Let's note that 1/17 is a repeating decimal. 

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

There are 16 repeating digits in the decimal. 

Dividing 4000 by 16, we get

\(4000/16 = 250\)

Since it is divisble, then we can conclude that the last digit of the repeating decimal is the correct answer. 

Thus, 7 is the final answer. 

Thanks! :)