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Input: x|x|=2x+1

Intepretation: \(x\left|x\right|=2x+1\)

Rewrite into alternative form:

\(x\sqrt{x^2}=2x+1\)

Simplify the square:

\(x^2=2x+1\)

Move the terms to the left:

\(x^2-2x-1=0\)

This function doesn't look like that it is factorizable, applying formulaic solution:

For \(ax^2+bx+c=0,\space x = {-b \pm \sqrt{b^2-4ac} \over 2a}\),

Plug \(a=1,\space b=-2,\space c=-1\)

\(x=\frac{2\pm \sqrt{4+4}}{2}\)

\(=\frac{2\pm \sqrt{8}}{2}\)

\(=\frac{2\pm 2\sqrt{2}}{2}\)

\(=1±\sqrt2\)

\(x=1+\sqrt{2}\ or\ 1-\sqrt{2}\)

What is a:   sqrt(4+sqrt(16+16a))+sqrt(1+sqrt(1+a))=6

(\(\sqrt{4+\sqrt{16+16a}}+\sqrt{1+\sqrt{1+a}}=6.\))

\(\begin{array}{|rcll|} \hline \sqrt{4+\sqrt{16+16a}}+\sqrt{1+\sqrt{1+a}} &=& 6 \\ \sqrt{4+\sqrt{16(1+a)}}+\sqrt{1+\sqrt{1+a}} &=& 6 \\ \sqrt{4+4\sqrt{1+a}}+\sqrt{1+\sqrt{1+a}} &=& 6 \\ \sqrt{4(1+\sqrt{1+a})}+\sqrt{1+\sqrt{1+a}} &=& 6 \\ 2\sqrt{1+\sqrt{1+a}}+\sqrt{1+\sqrt{1+a}} &=& 6 \\ 3\sqrt{1+\sqrt{1+a}} &=& 6 \quad & | \quad : 3 \\ \sqrt{1+\sqrt{1+a}} &=& 2 \quad & | \quad \text{square both sides} \\ 1+\sqrt{1+a} &=& 4 \quad & | \quad -1 \\ \sqrt{1+a} &=& 3 \quad & | \quad \text{square both sides} \\ 1+a &=& 9 \quad & | \quad -1 \\ \mathbf{ a } & \mathbf{=} & \mathbf{8} \\ \hline \end{array}\)

Proof:

\(\begin{array}{rcll} \sqrt{4+\sqrt{16+16a}}+\sqrt{1+\sqrt{1+a}} &\overset{?}{=}& 6 \qquad a = 8\\ \sqrt{4+\sqrt{16+16\cdot 8}}+\sqrt{1+\sqrt{1+8}} & \overset{?}{=} & 6 \\ \sqrt{4+\sqrt{144}}+\sqrt{1+3} & \overset{?}{=} & 6 \\ \sqrt{4+12}+\sqrt{4} & \overset{?}{=} & 6 \\ \sqrt{16}+2 & \overset{?}{=} & 6 \\ 4+2 & \overset{?}{=} & 6 \\ 6 & \overset{!}{=} & 6 \quad \checkmark\\ \end{array}\)

Input: What is a:   sqrt(4+sqrt(16+16a))+sqrt(1+sqrt(1+a))=6

Intepretation: Solve for \(a\) in \(\sqrt{4+\sqrt{16+16a}}+\sqrt{1+\sqrt{1+a}}=6\)

Simplify:

\(\sqrt{4+4\sqrt{1+a}}+\sqrt{1+\sqrt{1+a}}=6\)

We know that \(\sqrt{4+4\sqrt{1+a}}\) is \(\sqrt4=2\) times larger than \(\sqrt{1+\sqrt{1+a}}\)

Merge:

\(3\sqrt{1+\sqrt{1+a}}=6\)

Divide both sides by a factor of 3:

\(\sqrt{1+\sqrt{1+a}}=2\)

Since we know that \(\sqrt4=2\)

Therefore:

\(1+\sqrt{1+a}=4\)

\(\sqrt{1+a}=3\)

Since \(\sqrt9=3\)

\(1+a=9\)

\(a=8\)

Q.E.D.

(For one to solve this question, you just need to know the basic ideas of squares and powers (And some work) :P)

Input:nsolve(y+57+x+16+4x-5,x)

Intepretation:Solve \((y+57+x+16+4x-5=0)\) for \(x\)

\(y+57+x+16+4x-5=0\)

Merge the corresponding terms:

\(y+5x+68=0\)

Move \(y\) and \(68\) to the left:

\(5x=-y-68\)

Divide the entire function by a factor of 5:

\(x=-\frac{1}{5}y-\frac{68}{5}\)

\(=\frac{1}{5}\left(-y-68\right)\)

Q.E.D.

\(\)Input:$\log_{\sqrt{5}} 125\sqrt{5}$

Intepretation:

\(1.\log_{\sqrt{5}} 125\sqrt{5}\) (Scientific Calculator 2.0)

We want to know how many powers of \(\sqrt5\) equal to \(125\sqrt5\)

Looking at the numbers, we can know that the power is an integer, therefore:

\(1.\sqrt5^2=5\) \(,\space2.\sqrt5^3=5\sqrt5\)

\(3.\sqrt5^4=25\)\(,\space4.\sqrt5^5=25\sqrt5\)

\(5.\sqrt5^6=125\)\(,\space6.\sqrt5^7=125\sqrt5\)

Answer \(=7\)

Intepretation No.2:
\(2.\sqrt5\cdot125\sqrt5\) (Wolfram Computational Engine)

Cancel the two \(\sqrt5\) (\(\sqrt5^2=5\))

\(=5\cdot125\)

\(=625\)

Answer \(=625\)

Light year = 365 days x 24 hours x 60 minutes x 60 seconds x 186,282 miles/sec

                   = 5,874,589,152,000 miles in 1 light-year.

80 / 360 = 2/9

Area of circle = Pi x Radius^2

32Pi = 2/9 * Pi * r^2
Solve for r:
32 π = (2 π r^2)/9

32 π = (2 π r^2)/9 is equivalent to (2 π r^2)/9 = 32 π:
(2 π r^2)/9 = 32 π

Divide both sides by (2 π)/9:
r^2 = 144

Take the square root of both sides:
Answer: | r = 12  units

Well, technically it would be 11, but if you are just looking for -|-11|, it would be -11, but you asked for the absolute value of -|-11| and -|-11| = -11, so then the absolute value of that would look like |-11|, and the answer to that would be 11.

- l - 11 l  =

- 11

You would take off 3 pennies and the remaining 5 coins would have to be: 3 nickels + 2 dimes. So:

2 dimes = 20 cents

3 nickels = 15 cents

3 pennies = 3 cents.

Or, 8 coins = 38 cents.

Here's another method to prove condition ii (\(\overline{AC}\perp\overline{DB}\)). I will utilize a two-column proof:

x + 3/12  =  7/6 + (x + 6)/4

Simpliby 3/12 to 1/4:

x + 1/4  =  7/6 + (x + 6)/4

Multiply both sides by 12, the lowest common denominator:

12[x + 1/4]  =  12[7/6 + (x + 6)/4]

Use the distributive property:

12x + 3  =  14 + 3(x + 6)

Simplify:

12x + 3  =  14 + 3x + 18

Simplify:

12x + 3  =  3x + 32

Subtract 3x from both sides:

9x + 3  =  32

Subtract 3 from both sides:

9x  =  29

Divide both sides by 3:

x = 29/9

2  =  a*4

Divide both sides by 4:

2/4  =  a

a  =  1/2

log₂(4x + 3)  =  3

--->     4x + 3  =  2³

--->     4x + 3  =  8

--->           4x  =  5

--->             x  =  5/4

AD is congruent to CD (given) 

AB is congruent to CB (given)

DB is contruent to DB (identity)

Therefore, triangle(ADB) is congruent to triangle(DCB) (side-side-side)

and angle(ADB) is congruent to angle(CDB) (corresponding parts of congruent triangles are congruent)

Since angle(ADB) is congruent to angle(CDG), BD bisects angle(ADC).

AD is congruent to CD (given)

angle(ADB) is congruent to angleCDB) (proven above)

Therefore, triangle(ADE) is congruent to triangle(CDE) (side-angle-side)

and angle(AED) is congruent to angle(CED) (corresponding parts of congruent triangles are congruent)

Since angle(AED) and angle(CED) formf a linear pair (a straight line), angle(AED) and angle(CED) are right angles.

Thus, AC and DB are perpendicular.

The opposite vertices of a square are the endpoints of a diameter of that square.

Therefore, the angle between the x- and y-sides of the square is a right angle and x and y are the sides of a right triangle.

Since the radius of the circle is 5, its diameter is 10:

x² + y²  = 10².

Thank you soo much CPhil, but unfortunately that isn't the correct answer. I am taking a Probability and Counting class on AOPS like Mellie was....

how many type of fish species are there that have not been discoved yet?

HHHHEEEELLLLPPPP!!!!!!!!!! I've been stuck on this question forever.

THX SO MUCH!!!!!!!

Solve for x:
5 (x - 6) - 6 (x - 5) = 8

5 (x - 6) = 5 x - 30:
5 x - 30 - 6 (x - 5) = 8

-6 (x - 5) = 30 - 6 x:
-30 + 5 x + 30 - 6 x = 8

Grouping like terms, -30 + 5 x + 30 - 6 x = (30 - 30) + (5 x - 6 x):
(30 - 30) + (5 x - 6 x) = 8

5 x - 6 x = -x:
-x + (30 - 30) = 8

30 - 30 = 0:
-x = 8

Multiply both sides of -x = 8 by -1:
(-x)/(-1) = -8

(-1)/(-1) = 1:
Answer: | x = -8

No side can be < 2  or > 5

Here are the possibilities  for the sides :

 2 5 5     3 4 5      4 4 4

1.5E(1-1/8 E/D) E= 0.062, D= 0.5

Without parentheses, this is a little difficult to interpret....but....by order of operations we have

1.5 (.062) ( 1 - (1/8)(.062) / .5 )  =

.093 ( 1 - .00775 /.5 )  =

.093 ( 1 - .0155)  =

.093 ( .9845)

0.0915585