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Where is K located? 

Do you have a picture? I mean, you don't need a pic, but where is K?

I will give you a hint,

whenever an equation is perendicular, it has the NEGATIVE RECIPROCAL of the original slope.

This is an example problem:

y = 2x + 9 and passes through ( 4, 2)?

So new slope is 2 =>   -1/2

y = -1/2x + b

2 = -4/2 + b             (Plug in coordinates (4, 2)

4 = b

So the equation is  y = -1/2x + 4

\(y = 2x-13\\ y = -3x+92\\ 2x-13 = -3x + 92\\ 5x = 105\\ x = 21\)

\(3x+4y=7\\ 6x+4y=c-4y\\~\\ 6x+8y=c\\ \text{letting $c=14$ makes equation 2 twice equation 1 and thus the system has infinite solutions}\)

\(ab^4 = 12\\ a^5 b^5 = 7776\\ a^5b^{20} = 12^5\\ b^{15}=\dfrac{12^5}{7776} = 32\\ b^3 = 2\\ b= 2^{1/3}\\ 2^{4/3} a= 12\\ a = 3 \cdot 2^{2/3}\)

\(\text{Add the two equations}\\ 2a = 8b\\ b=\dfrac a 4\\ a + a\left(\dfrac a 4\right)^2 = 40\left(\dfrac a 4\right)\\ \dfrac{a^3}{16} =9a\\ a^2 = 144\\ a= \pm 12\\ b=\pm 3\\ \text{$a=0,b=0$ is also a solution} \)

3 2D vectors cannot be linearly independent.

By eye

a=2

b=-3

c=7

5/7

Thank you very much!! 

Don't simplify 10/75

First, do 1 - (10/75)

(75/75) - (10/75)   =    65/75

THEN simplify

13/15

Dang thats a lot of work! But a good problem solved.

Let's start by expanding both sides and combining like terms:

\(\begin{array}{rcl} \left(a+\frac{1}{b}\right)\left(b+\frac{1}{c}\right)\left(c+\frac{1}{a}\right)&=& \left(1+\frac{1}{a}\right)\left(1+\frac{1}{b}\right)\left(1+\frac{1}{c}\right)\\~\\ \left(ab+\frac{a}{c}+1+\frac{1}{bc}\right)\left( c+\frac{1}{a} \right)&=&\left(1+\frac{1}{b}+\frac{1}{a}+\frac{1}{ab} \right)\left( 1+\frac{1}{c}\right)\\~\\ abc+b+a+\frac{1}{c}+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{abc}&=& 1+\frac{1}{c}+\frac{1}{b}+\frac{1}{bc}+\frac{1}{a}+\frac{1}{ac}+\frac{1}{ab}+\frac{1}{abc}\\~\\ abc+b+a+{\color{gray}\frac{1}{c}}+c+{\color{gray}\frac{1}{a}}+{\color{gray}\frac{1}{b}}+{\color{gray}\frac{1}{abc}}&=& 1+{\color{gray}\frac{1}{c}}+{\color{gray}\frac{1}{b}}+\frac{1}{bc}+{\color{gray}\frac{1}{a}}+\frac{1}{ac}+\frac{1}{ab}+{\color{gray}\frac{1}{abc}}\\~\\ abc+b+a+c&=&1+\frac{1}{bc}+\frac{1}{ac}+\frac{1}{ab}\\~\\ abc+a+b+c&=&1+\frac{a}{abc}+\frac{b}{abc}+\frac{c}{abc}\\~\\ abc+a+b+c&=&1+\frac{a+b+c}{abc} \end{array}\)

At this point, let's substitute   x   in for   a + b + c   and   13   in for   abc    so we want to solve this for  x:

\(\begin{array}{ccc} 13+x&=&1+\frac{x}{13}\\~\\ 169+13x&=&13+x\\~\\ 13x-x&=&13-169\\~\\ 12x&=&-156\\~\\ x&=&-13 \end{array}\)_

You can also do some modulus, but Im too lazy for dat

Let m be a multiple of 5.

n is one more than a multiple of 5, so n=m+1

3n=3(m+1)=3m+3

3m will always be a multiple of 5 because m is a multiple of 5, so the remainder is 3.

Instead of knowing the constants and solving for a point, we are given two points and asked to solve for the constants.

First plug in the two points to the equation.

(2,3)=(x,y) plugged into y=x^2+bx+c

3=2^2+2b+c

3=4+2b+c

subtract 4 from each side

-1=2b+c

(4,3)

3=4^2+4b+c

3=16+4b+c

subtract 16 from each side

-13=4b+c

Now this problem is only a matter of solving a system of equations:

-13=4b+c

-1=2b+c

Subtract the second one from the first one to eliminate c:

-12=2b

divide by 2

b= -6

Substitute -6 in for b to get c:

-1=2(-6)+c

-1=-12+c

add 12 to each side

11=c

MAD stands for mean absolute deviation.

To find the mad, first find the mean of the data set. Next find the absolute difference between the mean and all of the pieces of data. Then find the mean of the differences to get the mad.

https://www.khanacademy.org/math/cc-sixth-grade-math/cc-6th-data-statistics/cc-6-mad/v/mean-absolute-deviation

It's pretty obvious that the remainder is 3 but what is the math behind it

Solution:

\(\small \text{There are } \dbinom{16}{4} = 1820 \text{ sets of 4-cards. Four of these sets contain exactly one (1) card with a one.}\\ \text {Probability: } \dfrac{4}{1820} = \dfrac{1}{455}\\\)

GA

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I’m leaving this train wreck of a solution here as a sign post to slow down the trains of thought, before changing tracks. 

I initially solved this using (4^4) as the number of arrangements where the “1 cards” could occupy any row in the four columns of four rows (this is easy to see). Then dividing this by the total number of combinations,(4⁴) / (nCr(16,4)), gives (64/445) = 0.1406593406593407, the probability.  

For some reason, I thought this was too high of a probability.... 

So, what did I do to “fix” it? I’m not sure ... took the fourth root before dividing it (maybe –IDFK). After changing tracks, I didn’t realize the train actually jumped the tracks until the caboose ended up where the engine should be.  This is a simple, high school level probability problem that my cat could solve when buzzed on catnip, yet I turned it into a bloody train wreck. 

I am a much better troll than I am a mathematician.

Melody’s solution is defined and nuanced at every step. It’s even better than my cat could do.

GA

This is a circle.....

"complete the square"  for x and y to get

(x+3)^2 + (y+4)^2 =5^2       so the center is -3,-4 and the radius is 5    Area of a circle = pi r^2   = 25 pi

Desmos graphical solution:

deleted

thx i  was overthinking it, so simple

How did you get that?

I don't understand how to do it.

I believe the answer is  0 .

That IS the graph of a tangent function....specifically  tan (x/3)

The period is the distance on the x-axis before the graph repeats itself

3pi/2 - - 3pi/2 = 3pi

Here is a desmos graph of tan (x/3)

https://www.desmos.com/calculator/ny77s8o78x

For comparison.....here is a graph of tan x

https://www.desmos.com/calculator/t4yhf9vfzq

\(a_n=n\)

is the formula

Its like a is 0  so n=n 

15=15 

16=16 etc..

 The formula a_n=n says that "a" which is a term, n identify which term it is, so first term we would write it  as a_1 = As the sequence says 

here we see that a_1=1 and so on for a_15=15 

so we can make a general formula (Explicit) which is a_n=n for n is any number.