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Latex is a mathematical language in which you can write to make your proof or work easier to read and understand.

Let's call the side length of the larger square  " s "  .

Then...

diameter of the larger circle  =  s  

Let's draw a diameter of the larger circle from the corner of the smaller square to the opposite corner of the smaller square. This creates a 45 - 45 - 90 triangle, where the side across from the 90° angle is  "s"  . So....

side across from the  45°  angle  =   \(\frac{s}{\sqrt2}\)

diameter of the smaller circle       =  \(\frac{s}{\sqrt2}\)

radius of the smaller circle            =  \(\frac12\,*\,\text{diameter}=\frac12\,*\,\frac{s}{\sqrt2}=\frac{s}{2\sqrt2}\)

\(\frac{\text{area of smaller circle}}{\text{area of larger square}}=\frac{\pi(\frac{s}{2\sqrt2})^2}{s^2}=\frac{\pi(\frac{s^2}{4*2})}{s^2}=\frac{\pi s^2}{8}\,*\,\frac1{s^2}=\frac{\pi}{8}\)

a  = sqrt [ 4 + sqrt (4 + a) ]     square both sides

a^2  = 4 + sqrt ( 4 + a)

a^2 - 4 = sqrt (4 + a)       square both sides again

a^4 - 8a^2 + 16  = 4 + a

a^4 - 8a^2 - a + 12  = 0

There are 4 possible real solutions to this 

The only correct one  is  ...    a  = [ 1 + sqrt (17 ] / 2

Likewise...setting each of the other equations up as 4th power polynomials by squaring each side twice produces

b = [ 1 + sqrt (13 ) ] / 2

c = [-1 + sqrt(17) ] / 2

d =  [ -1 + sqrt (13) ] / 2

So....abcd  =  acbd  =

[ 1 + sqrt (17)] / 2  *   [-1 + sqrt(17) ] /2  *   [ 1 + sqrt (13 ) ] / 2  * [ -1 + sqrt (13) ] / 2 =

[ -1 + 17 ] / 4    *   [ -1 + 13] / 4  =

[ 16 / 4 ] * [ 12 / 4 ]  =

4  *  3   =  

 12  

(-6) over (-2) = -6*(Error: Unknown Function 'over')

2  + 2/9= (-4/5 )m

2 + 2/9  =  20/9      so we have

20 / 9  =  ( - 4 / 5 )  m         multiply both sides by  -5 / 4

( 20 / 9 ) ( - 5 / 4)  = m

-100 / 36  = m            reduce

-25 / 9   = m

(x + 2) / 15 =  (x + 1 ) / 5         cross-multiply

5 ( x + 2)  = 15 (x + 1)    simplify

5x + 10  = 15x  + 15       subtract 15, 5x from both sides

-5 = 10x                   divide both sides by 10

-5 / 10  = x  =  -1/2

 If s(x) = 2x^2 + 3x - 4, and t(x) = x + 4 

s(x) * t (x )  =       [ 2x^2 + 3x - 4 ]  [ x + 4 ]     =    [ 2x^3 + 3x^2 - 4x  + 8x^2 + 12x - 16 ]  =

 [ 2x^3 + 11x^2  + 8x  - 16 ]

It looks as though f(x) = g (x ) at  (-2, -1)

[x^2 + x - 6] / [x^2 - 4] *  [x^2 - 9] / [x^2 + 6x + 9]     factor numerators and denominators

[ (x + 3) (x - 2)] / [ (x + 2) ( x - 2) ]  *  [ ( x + 3) ( x - 3) ] / [ (x + 3) ( x + 3) ]

Simplifies to

[ ( x + 3) ] / [ ( x + 2) ]  *  [ ( x - 3) ] / [ ( x + 3) ]    =

[ ( x + 3) ] / [ ( x + 3) ] * [ x - 3) ] /  [ ( x + 2) ]  =

[ ( x - 3) ] / [ ( x + 2) ]

[x ^3 - 2x^2 + 3  ] / [ x -2 ]

Because the polynomial in the  numerator is of a greater degree than the polynomial in the denominator, there is no horizontal asymptote

x^2 + 5x - 14                                  [ ( x + 7) (x - 2) ]                         [ x - 2 ]

__________        factors as           _____________      =                ______    

x^2 + 8x + 7                                   [ (x + 7) ( x + 1) ]                        [ x + 1 ]    

To solve this problem, first find the slope of the first two points.

\(m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\)

\(m=slope\)

\({y}_{2}=\)y-point in second point

\({y}_{1}=\)y-point in first point

\({x}_{2}=\)x-point in second point

\({x}_{1}=\)x-point in first point

\({y}_{2}=3\)

\({y}_{1}=4\)

\({x}_{2}=6\)

\({x}_{1}=2\)

\(m=\frac{3-4}{6-2}\)

\(m=\frac{-1}{6-2}\)

\(m=\frac{-1}{4}\)

\(m=-\frac{1}{4}\)

Next, do the same for the second and third points and solve for c.

\(m=-\frac{1}{4}\)

\({y}_{2}=c\)

\({y}_{1}=3\)

\({x}_{2}=-5\)

\({x}_{1}=6\)

\(-\frac{1}{4}=\frac{c-3}{-5-6}\)

\(-\frac{1}{4}=\frac{c-3}{-11}\)

\(-\frac{1}{4}\times-11=\frac{c-3}{-11}\times-11\)

\(\frac{11}{4}=\frac{c-3}{-11}\times-11\)

\(\frac{11}{4}=c-3\)

\(\frac{11}{4}+3=c-3+3\)

\(\frac{11}{4}+\frac{12}{4}=c-3+3\)

\(\frac{11+12}{4}=c-3+3\)

\(\frac{23}{4}=c-3+3\)

\(5.75=c-3+3\)

\(5.75=c-0\)

\(5.75=c\)

\(c=5.75\)

Here's another method that works, too. Let's use the zero-product theorem. If haven't heard of the theorem, then you have probably used it before, unbeknown to you. With this theorem, we can make the following equation with the roots of \(\pm2\)

\((x+2)(x-2)=0\)

The zero-product theorem states that at least one (or both) factors must be equal to zero for the left-hand side to equal 0. I have made an equation that contains both of the desired roots. Since we know that this equation has the desired roots and \(x^2+b-10=0\) has the same desired roots, we can conclude that \((x+2)(x-2)=x^2+b-10\). Now, solve for b:

This is simply another method of solving. 
 

If your question is to find the first derivative:

f(x) = sin(2x)     --->     f'(x)  = 2·cos(2x)                             (using the chain rule)

Given the formula:  h  =  -16t² + 100t + 70  and given the value:  h = 30,  what is the value of t?

30  =  -16t² + 100t + 70

  0  =  -16t² + 100t + 40                                                       subtract 30 from both sides

  0  =  -4t² + 25t + 10                                                           divide both sides by 4

a = -4     b = 25     c = 10                                                     use the quadratic formula

t  =  [ -b +/- sqrt( b² - 4·a·c ) ] / (2·a)

t  =  [ -25 +/- sqrt( (25)² - 4·(-4)·(10 ) ] / (2·-4)

t  =  [ -25 +/- sqrt( 625 + 160) ] / (-8)

t  =  [ -25 +/- sqrt( 785) ] / (-8)

t = -0,377     or     t = 6.627

Since t must be positive, t = 6.6

That was a creative way of making the symbol for pi! For future reference, use /pi in Latex to get the desired symbol. It looks like this:

\(\pi\)

 x + 2  =  5x/2 - (8x - 5)/4

Multiply each side by 4:

4[x + 2]  =  4[5x/2] - 4[ (8x - 5)/4 ]

Using the distributive principle and simplifying fractions:

4x + 8  =  2(5x) - (8x - 5)

4x + 8  =  10x - 8x + 5

4x + 8  =  2x + 5

Subtract 2x from both sides:

2x + 8  =  5

Subtract 8 from both sides:

2x  =  -3

Divide both sides by 2:

x  =  -3/2

a  = sqrt [ 4 + sqrt (4 + a) ]     square both sides

a^2  = 4 + sqrt ( 4 + a)

a^2 - 4 = sqrt (4 + a)       square both sides again

a^4 - 8a^2 + 16  = 4 + a

a^4 - 8a^2 - a + 12  = 0

There are 4 possible real solutions to this 

The only correct one  is  ...    a  = [ 1 + sqrt (17 ] / 2

Likewise...setting each of the other equations up as 4th power polynomials by squaring each side twice produces

b = [ 1 + sqrt (13 ) ] / 2

c = [-1 + sqrt(17) ] / 2

d =  [ -1 + sqrt (13) ] / 2

So....abcd  =  acbd  =

[ 1 + sqrt (17)] / 2  *   [-1 + sqrt(17) ] /2  *   [ 1 + sqrt (13 ) ] / 2  * [ -1 + sqrt (13) ] / 2 =

[ -1 + 17 ] / 4    *   [ -1 + 13] / 4  =

[ 16 / 4 ] * [ 12 / 4 ]  =

4  *  3   =  

 12  

yea, thats it thanks

The guest is correct......a circle will give us the greatest area

The area if the whole sector  is just (1/4)  that of the circle (since BOA is a right angle )

So...area of the sector  =  (1/4) pi * radius^2   =  (1/4) pi * 8^2  =  (1/4) pi * 64  =  (1/4) * 64 * pi  = 16 pi   units'^2

The area of the triangle AOB  is just   (1/2) radius^2  =  (1/2) (8)^2 =  (1/2) 64   = 32 units^2

So....the purple area is just the area of the sector  - area of the triangle AOB  = 

[ 16 pi - 32]  units^2  ≈  18.26 units^2  

If

\(\frac{a+b}{c+d}=\frac{16}{15}\)

then what is a,b,c and d?

Or just, what is abcd=?

Another possibly dumb question my classmate made and we tried to solve.

\(\frac{9+7}{10+5}=\frac{16}{15}\)

\(a=9,\ b=7,\ c=10,\ d=5\)

\( abcd=9\times 7\times 10\times 5=3150\)

Was that asked?

!

This is a combinations problem

There are 4  seventh graders and we want to choose one of them......

The number of ways this can be done  =  C (4,1) = 4      (1)

There are  six eighth graders and we want to choose any three of them

The number of ways this can be done  = C(8,3)  = 56      (2)

So....the total possible committees  =  (1) * (2)   =  (4) * (56)  =  224 

I'm assuming that x₁  and x₂  are the roots

If the roots are  2 and -2, the we have that, using either one of them.....

(2)^2  + b - 10  = 0

4 + b - 10  = 0

-6 + b  = 0

b = 6

Then our function becomes

x^2 + 6 -  10  =0

x^2 - 4  = 0

Check to see that x = 2 and x  = - 2 make this  = 0

Hi Max: I'm sure that there is a SHORTER solution for "a", but my "Mathematica 11" gives this ridiculously long answer. Sorry about that.

Solve for a:
a = sqrt(sqrt(a + 4) + 4)

a = sqrt(sqrt(a + 4) + 4) is equivalent to sqrt(sqrt(a + 4) + 4) = a:
sqrt(sqrt(a + 4) + 4) = a

Raise both sides to the power of two:
sqrt(a + 4) + 4 = a^2

Subtract 4 from both sides:
sqrt(a + 4) = a^2 - 4

Raise both sides to the power of two:
a + 4 = (a^2 - 4)^2

Expand out terms of the right hand side:
a + 4 = a^4 - 8 a^2 + 16

Subtract a^4 - 8 a^2 + 16 from both sides:
-a^4 + 8 a^2 + a - 12 = 0

The left hand side factors into a product with three terms:
-(a^2 - a - 4) (a^2 + a - 3) = 0

Multiply both sides by -1:
(a^2 - a - 4) (a^2 + a - 3) = 0

Split into two equations:
a^2 - a - 4 = 0 or a^2 + a - 3 = 0

Add 4 to both sides:
a^2 - a = 4 or a^2 + a - 3 = 0

Add 1/4 to both sides:
a^2 - a + 1/4 = 17/4 or a^2 + a - 3 = 0

Write the left hand side as a square:
(a - 1/2)^2 = 17/4 or a^2 + a - 3 = 0

Take the square root of both sides:
a - 1/2 = sqrt(17)/2 or a - 1/2 = -sqrt(17)/2 or a^2 + a - 3 = 0

Add 1/2 to both sides:
a = 1/2 + sqrt(17)/2 or a - 1/2 = -sqrt(17)/2 or a^2 + a - 3 = 0

Add 1/2 to both sides:
a = 1/2 + sqrt(17)/2 or a = 1/2 - sqrt(17)/2 or a^2 + a - 3 = 0

Add 3 to both sides:
a = 1/2 + sqrt(17)/2 or a = 1/2 - sqrt(17)/2 or a^2 + a = 3

Add 1/4 to both sides:
a = 1/2 + sqrt(17)/2 or a = 1/2 - sqrt(17)/2 or a^2 + a + 1/4 = 13/4

Write the left hand side as a square:
a = 1/2 + sqrt(17)/2 or a = 1/2 - sqrt(17)/2 or (a + 1/2)^2 = 13/4

Take the square root of both sides:
a = 1/2 + sqrt(17)/2 or a = 1/2 - sqrt(17)/2 or a + 1/2 = sqrt(13)/2 or a + 1/2 = -sqrt(13)/2

Subtract 1/2 from both sides:
a = 1/2 + sqrt(17)/2 or a = 1/2 - sqrt(17)/2 or a = sqrt(13)/2 - 1/2 or a + 1/2 = -sqrt(13)/2

Subtract 1/2 from both sides:
a = 1/2 + sqrt(17)/2 or a = 1/2 - sqrt(17)/2 or a = sqrt(13)/2 - 1/2 or a = -1/2 - sqrt(13)/2

a ⇒ -1/2 - sqrt(13)/2 = 1/2 (-1 - sqrt(13)) ≈ -2.30278
sqrt(sqrt(a + 4) + 4) ⇒ sqrt(sqrt((-1/2 - (sqrt(13))/(2)) + 4) + 4) = sqrt(sqrt(7/2 - (sqrt(13))/(2)) + 4) ≈ 2.30278:
So this solution is incorrect

a ⇒ sqrt(13)/2 - 1/2 = 1/2 (sqrt(13) - 1) ≈ 1.30278
sqrt(sqrt(a + 4) + 4) ⇒ sqrt(sqrt(((sqrt(13))/(2) - 1/2) + 4) + 4) = sqrt(sqrt(7/2 + (sqrt(13))/(2)) + 4) ≈ 2.51053:
So this solution is incorrect

a ⇒ 1/2 - sqrt(17)/2 = 1/2 (1 - sqrt(17)) ≈ -1.56155
sqrt(sqrt(a + 4) + 4) ⇒ sqrt(sqrt((1/2 - (sqrt(17))/(2)) + 4) + 4) = sqrt(sqrt(9/2 - (sqrt(17))/(2)) + 4) ≈ 2.35829:
So this solution is incorrect

a ⇒ 1/2 + sqrt(17)/2 = 1/2 (1 + sqrt(17)) ≈ 2.56155
sqrt(sqrt(a + 4) + 4) ⇒ sqrt(sqrt((1/2 + (sqrt(17))/(2)) + 4) + 4) = sqrt(sqrt((sqrt(17) + 9)/(2)) + 4) ≈ 2.56155:
So this solution is correct

The solution is:
Answer: | a = 1/2 + sqrt(17)/2 =2.5615528........

b = b = 1/2 (1 + sqrt(13)) = 2.30277564......

c=c = 1/2 (sqrt(17) - 1) = 1.561552813.....

d =d = 1/2 (sqrt(13) - 1) = 1.30277564........

abcd = ~ 12  The product of these 4 variables certainly does not equal 48 but 12 !!!!.

\(A\ decimeter\ (dm)\ is\ one\ tenth\ of\ a\ meter.\\ The\ diameter\ of\ Earth\ is\ 1.3 \times 10^4 km.\\ What\ is\ the\ diameter\ expressed\ in\ decimeters?\)

\(1.3 × 10^{–5} dm\\ 1.3 × 10^{11} dm\\ \color{blue}1.3 × 10^8 dm\\ 1.3 × 10^7 dm\)

\(The\ diameter\ of\ Earth\ is\ 1.3 \times 10^4\ km\times\frac{10^4dm}{km}=1.3\times10^8\ dm\)

  !

Identify the value that is NOT in the range of the piecewise function.

Let's look at each answer choice.

A) -2

There is a green point where  y = -2  ...so  -2  is in the range of the function.

B) 1

There is a blue point where  y = 1  ...so  1  is in the range of the function.

C) 2

There is a blue point where  y = 2  ...so  2  is in the range of the function.

D) 5

There is not a point where  y = 5  ...so  5  is not in the range of the function.

If the graph is that of  f(x) , name a point that lies on  f^-1(x)  .

If a point on  f(x)  is  (a, b)  , then a point on  f^-1(x)  will be  (b, a)  .

At the vertex of this graph, it looks like there is the point  (1, -3)  .

So, a point on inverse of this function will be  (-3, 1)  .

Simplify the following expression as far as possible:

xy+ 3x²y+ 1 + 2xy²+yx²+ 2 +yx

\(\begin{array}{|rcll|} \hline && xy+ 3x^2y+ 1 + 2xy^2+yx^2+ 2 +yx \\ &=& 3+2xy+ 4x^2y+ 2xy^2 \\ &=& 3+2xy+ 2xy(2x+ y) \\ &=& 3+2xy(1+2x+ y) \\ \hline \end{array}\)