All Answers 358187 Answers

hectictar: I just programmed the equation into my computer and it came back with the answer as:

x =2.16293888805 in less a second. It used Newton's method to home in on the answer. Perhaps, it is much easier and faster to solve such equations using this Newton's method. 

y + vv - ( v + y * y) / v * v

Replace every " y " with  15  , and replace every " v " with  5  .

15 + (5)(5) - (5 + 15 * 15) / 5 * 5

Work inside the parenthesees first. Within the parenthesees, multiplication comes before addition.

15 + (5)(5) - (5 + 225) / 5 * 5

15 + (5)(5) - (230) / 5 * 5

Work on the multiplication and division next.

15 + 25 - 46 * 5

15 + 25 - 230

Finally addition and subtraction.

40 - 230

- 190

I don't think there is a shorter  " exact "  solution than that.... because in order to get x all by itself on one side of the equation, you must go through the process of solving a cubic equation.

Ugh, these jokes.

"Nothing, they just waved"

"Did you sea what I did there?"

"Don't be such a beach"

Here is a "ridiculous" solution from Mathematica 11, IF anybody cares to go through it. There surely MUST be a shorter solution!!.

Solve for x:
(3 log(x - 2))/(log(10)) = (log(2 x))/(log(10)) - 3

Subtract (log(2 x))/(log(10)) - 3 from both sides:
3 + (3 log(x - 2))/(log(10)) - (log(2 x))/(log(10)) = 0

Bring 3 + (3 log(x - 2))/(log(10)) - (log(2 x))/(log(10)) together using the common denominator log(10):
(3 log(10) + 3 log(x - 2) - log(2 x))/(log(10)) = 0

Multiply both sides by log(10):
3 log(10) + 3 log(x - 2) - log(2 x) = 0

3 log(10) + 3 log(x - 2) - log(2 x) = log(1000) + log((x - 2)^3) + log(1/(2 x)) = log((500 (x - 2)^3)/x):
log((500 (x - 2)^3)/x) = 0

Cancel logarithms by taking exp of both sides:
(500 (x - 2)^3)/x = 1

Multiply both sides by x:
500 (x - 2)^3 = x

Subtract x from both sides:
500 (x - 2)^3 - x = 0

Expand out terms of the left hand side:
500 x^3 - 3000 x^2 + 5999 x - 4000 = 0

Eliminate the quadratic term by substituting y = x - 2:
-4000 + 5999 (y + 2) - 3000 (y + 2)^2 + 500 (y + 2)^3 = 0

Expand out terms of the left hand side:
500 y^3 - y - 2 = 0

Divide both sides by 500:
y^3 - y/500 - 1/250 = 0

Change coordinates by substituting y = z + λ/z, where λ is a constant value that will be determined later:
-1/250 + 1/500 (-z - λ/z) + (z + λ/z)^3 = 0

Multiply both sides by z^3 and collect in terms of z:
z^6 + z^4 (3 λ - 1/500) - (z^3)/(250) + z^2 (3 λ^2 - (λ)/(500)) + λ^3 = 0

Substitute λ = 1/1500 and then u = z^3, yielding a quadratic equation in the variable u:
u^2 - u/250 + 1/3375000000 = 0

Find the positive solution to the quadratic equation:
u = (450 + sqrt(202485))/225000

Substitute back for u = z^3:
z^3 = (450 + sqrt(202485))/225000

Taking cube roots gives (450 + sqrt(202485))^(1/3)/(10 15^(2/3)) times the third roots of unity:
z = (450 + sqrt(202485))^(1/3)/(10 15^(2/3)) or z = -(-450 - sqrt(202485))^(1/3)/(10 15^(2/3)) or z = ((-1)^(2/3) (450 + sqrt(202485))^(1/3))/(10 15^(2/3))

Substitute each value of z into y = z + 1/(1500 z):
y = (sqrt(202485) + 450)^(1/3)/(10 15^(2/3)) + 1/(10 (15 (sqrt(202485) + 450))^(1/3)) or y = (-1)^(2/3)/(10 (15 (sqrt(202485) + 450))^(1/3)) - (-sqrt(202485) - 450)^(1/3)/(10 15^(2/3)) or y = ((-1)^(2/3) (sqrt(202485) + 450)^(1/3))/(10 15^(2/3)) - 1/10 ((-1)/(15 (sqrt(202485) + 450)))^(1/3)

Bring each solution to a common denominator and simplify:
y = ((sqrt(202485) + 450)^(2/3) + 15^(1/3))/(10 15^(2/3) (450 + sqrt(202485))^(1/3)) or y = ((-15)^(2/3) - (-15)^(1/3) (sqrt(202485) + 450)^(2/3))/(150 (450 + sqrt(202485))^(1/3)) or y = -((sqrt(202485) - 450)^(1/3) - (-1)^(2/3) (sqrt(202485) + 450)^(1/3))/(10 15^(2/3))

Substitute back for x = y + 2:
x = (15^(1/3) + (sqrt(202485) + 450)^(2/3))/(10 15^(2/3) (sqrt(202485) + 450)^(1/3)) + 2 or x = ((-15)^(2/3) - (-15)^(1/3) (sqrt(202485) + 450)^(2/3))/(150 (sqrt(202485) + 450)^(1/3)) + 2 or x = 2 - ((sqrt(202485) - 450)^(1/3) - (-1)^(2/3) (450 + sqrt(202485))^(1/3))/(10 15^(2/3))

(3 log(x - 2))/(log(10)) ≈ -2.41495 + 2.759 i
(log(2 x))/(log(10)) - 3 ≈ -2.41495 + 0.030247 i:
So this solution is incorrect

(3 log(x - 2))/(log(10)) ≈ -2.36393
(log(2 x))/(log(10)) - 3 ≈ -2.36393:
So this solution is correct

(3 log(x - 2))/(log(10)) ≈ -2.41495 - 2.759 i
(log(2 x))/(log(10)) - 3 ≈ -2.41495 - 0.030247 i:
So this solution is incorrect

The solution is:
Answer: | x = (15^(1/3) + (sqrt(202485) + 450)^(2/3))/(10 15^(2/3) (sqrt(202485) + 450)^(1/3)) + 2 = 2.1629388880535156

@MaxWong Thank you for explaining and simplifying the equation so I could understand it better! :D

Let x = the thing that you typed <-- too lazy XD

\(x =\cfrac{1}{1+\dfrac{1}{2+x}}\\ x + \dfrac{x}{2+x}=1\\\ x(x + 2) + x = x + 2\\ x^2 + 3x - x - 2 = 0\\ x^2 + 2x - 2 = 0\\ x = \dfrac{-(2)\pm\sqrt{2^2 - 4(1)(-2)}}{2(1)}\\ x =\dfrac{-2 \pm 2\sqrt{3}}{2}=-1\pm\sqrt{3}\\ x = \sqrt3 - 1\text{ or }x = -1-\sqrt3\text{(rej.) <-- Do you even think it's a negative number?}\\ x = \sqrt3 - 1\)

Search it on google, not here!!

\(x\cdot(n+x+\dfrac{n}{n})+ xn - n\\ = 3\cdot(5+3+5\div 5)+3\times 5 - 5\\ =3\cdot 9 + 15 - 5\\ = 27 + 15 - 5\\ = 42 - 5\\ = 37\)

We can use the distance formula on this problem:

distance between  (x₁ , y₁)  and  (x₂ , y₂)   =   \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\)

____________________________________________________________

ST = SO

distance between point S and point T = distance between point S and point O

distance between  (0,p)  and  (6, -2)   =   distance between  (0,p)  and  (0,0)

\(\sqrt{(0-6)^2+(p--2)^2}=\sqrt{(0-0)^2+(p-0)^2}\)           Square both sides of this equation.

(0 - 6)² + (p - -2)²   =   (0-0)² + (p - 0)²

(-6)² + (p + 2)²   =   (0)² + (p)²

36 + p² + 4p + 4   =   p²

40 + 4p   =   0

4p   =   -40

p   = -10

Evaluate x (n + x + n/n) + x n - n where x = 3 and n = 5:
x (n + x + n/n) + x n - n = 3 (5 + 3 + 5/5) + 3×5 - 5

5/5 = 1:
3 (5 + 3 + 1) + 3×5 - 5

5 + 3 + 1 = 9:
3×9 + 3×5 - 5

3×9 = 27:
27 + 3×5 - 5

3×5 = 15:
27 + 15 - 5
Answer: | 37

Here are some ways to calculate pi.

\((2*(\frac{1}{2})!)^2=\pi\)

\(\frac{4}{1}-\frac{4}{3}+\frac{4}{5}-\frac{4}{7}+\frac{4}{9}-\frac{4}{11}+\frac{4}{13}-\frac{4}{15}\hspace{1mm}...=\pi\)

Cool, huh?

I don't know what it is!

Hurray to you!

so???

1/4+1/2=1/4+2/4=3/4

That's me... Mr. DetailedAnswer ... 

Well....thanks, BP.....but.....you might want to save any "worship" for something else.....what I did was nothing  "miraculous"......LOL!!!!

Thanks for that detailed answer, X²...!!!

Okay, I can't seem to fix that typo, \((x<\frac{10}{7},\frac{16}{7}. I've tried and tried and tried, but it simply never saves my correction afterward. It's annoying, so annoying. Anyway, it's supposed to be:

(x<10/7,16/7 [insert less than symbol] y [insert less than and equal to] 16/7

THANK YOU MATH GOD. YOU SAVED MY LIFE. I WILL WORSHIP YOU EVERYDAY.

I assume you want to know each side length ???

If I'm seeing this correctly,   we have

[3x + 4]  + [2x + 3]  +  [ 2x  - 9  ]   =  75   simplify

7x  - 2  =  75      add 2 to both sides

7x   =  77            divide by 7

x  =  11

So,  one side  =   3(11) + 4   =   37

One side = 2(11) + 3  =  25

And one side  = 2(11) - 9  =  13

You can also solve this system algebraically, too, but it requires some logical thinking. You solve it similarly as that of a system of equations. Here are your inequalities:

1. \(2x+4y\leq12\)

2. \(3x-y<2\)

I'll use substitution. I'll solve for y on both equations. I'll solve for y on the inequality \(2x+4y\leq12\):

Next, isolate y in the inequality 2, \(3x-y<2\):

Our inequalities have changed to from their original to being solved for y:

1. \(y\leq-\frac{1}{2}x+3\)
2. \(y>3x-2 \)

This is where the logical thinking comes to play.If 3x-2 is less than y and \(-\frac{1}{2}x+3\) is equal to and greater than y, then \(3x-2<-\frac{1}{2}+3\)! Now that there is one variable in this inequality, we can solve for x.

Yet again, it requires a bit of logic again. Plug in 10/7 into both inequalities! Yes, you must plug it into both inequalities! You'll see why after the calculations are made. First, I'll plug x into the first and second equation

Great! Let's plug in x for the second equation and see what we get:

Now that we know our x- and y-values, let's write them out as a solution set:

\(((x<\frac{10}{7},\frac{16}{7}

I guess you could stop here and move on to the next problem, but do you notice how y is really equal to every number? Currently, the solution set is saying that y is greater than 16/7, equal to 16/7 and less than 16/7. That's a verbose way of saying that y can be any of the real numbers. I'll write it out for you. Sorry, I had a hard time making the real-number symbol in LaTeX:
 

\((x\in{\rm I\!R}| x<\frac{16}{7},y\in{\rm I\!R})\)

You are done now because you have identified the possible values for x and y! You can verify this solution set by graphing. Luckily, Cphill has already done that! You can check it out, if you'd like.

One last note before you go!

I spent around 3 hours in total (with a slight respite in between) to generate this response. I truly attempted to provide a complete answer with comprehensible explanations. I hope it helped! Also, if anyone sees a typo, please tell me. I'm 95% confident that I missed one or two.

"C" is the correct one.....we're substituting "10" for "t"

 h =  (1/2) (10)^2  =

        (1/2)  (100) = 

         50