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The answer is 22/9.

The answer is 65/72.

First, let's write this in factored form. 

A quadratic whose roots are 5 and -5 can be written in the form (x-5)(x+5). 

Expanding, we have \((x-5)(x+5) = x^2-25\). 

This means that the first blank, the coefficient for x, is 0. 

The other blank, the constant, is -25. 

We have a final answer of \(x^2+0x+25 \) or just \(x^2-25\)

Thanks! :)

I'm not exactly sure if I'm doing correctly, but I'll try my best. 

First off, let's note that \(a^x \cdot a^y = a^{x+y}\). 

Using this logic, we would have \(q^{\frac{5}{3}}p^{\frac{5}{4}}r^{\frac{31}{12}}\). 

In order to not have fractional exponents, we would have to put these into radical form. 

Note that \(a^{x/y} = \sqrt[y]{a^x}\).

We would get \(\sqrt[3]{q^5} \cdot \sqrt[4]{p^5} \cdot \sqrt[12]{r^{31}} \).

I hope I answered your question!

Thanks! :) 

3. As I said in question #2, we have \((x – h)^2 + (y – k)^2 = r^2\) where r is the radius. 

This means we have \(r^2 =169\), which just gives us \(r = \sqrt{169} = 13\). So the radius is 13. 

Note, the radius can't be -13 because the length or radius can't be negative!

Thanks! :)

2. The equation of a circle is \((x – h)^2 + (y – k)^2 = r^2\) where \((h,k)\) is the center of the circle and \(r\) is the radius. 

Now, we can just plug in the information we were already given in the equestion. 

If the center is (11, -9), we have \(h=11\) and \(k=-9\) and radius of 12. 

We have \((x-11)^2+(y+9)^2 = 144\). 

Thanks! :)

Simplify as

x^2 - 6x + 13

uv  =  13

2uv  = 26

u + v =  6       square both sides

u^2 + 2uv + v^2  = 36

u^2 + 26 + v^2  = 36

u^2 + v^2  = 10

u/v^2  + v/u^2  =  [ u^3 + v^3 ] / [ uv]^2

u^3 + v^3  = ( u + v) ( u^2 + v^2 - uv)  = (6)(10 - 13)  =  -18

[uv]^2 = [ 13]^2   = 169

u/v^2  + v/u^2  =  [ u^3 + v^3 ] / [ uv]^2 =  -18 / 169

Simplify as

10x^2 + 10x - 1

ab = -1/10

2ab = -2/10  = -1/5

a + b  =  -1      square both sides

a^2 + 2ab + b^2  = 1

a^2 - 1/5 + b^2  = 1

a^2 + b^2  = 1 + 1/5  =  6/5

(a -4) / ( b -4)  +  ( b -4) / (a -4)  =     [  (a - 4)^2  + ( b -4)^2] ]  ' [ (a -4) ( b-4) ]  =

[ a^2 - 8a +16  + b^2 - 8b + 16 ]  /  [ ab - 4(a + b) +16 ]   =

[  (a^2 + b^2) - 8(a + b) + 32 ]  / [ ab - 4 (a + b) + 16 ]  =

[ 6/5 - 8 (-1) + 32 ]  [ -1/10 - 4 (-1) + 16 ]  = 

[ 6/5 + 40 ]  / [ 20 - 1/10] =   20497 / 25

A²  = [1/sqrt 2    -1/sqrt 2]    *   [  1/sqrt 2   -1/sqrt2 ]   =   [ 0  -1 ]

         [ 1/sqrt 2    1/sqrt 2]         [   1/sqrt 2   1/sqrt2]         [  1  0 ]

A⁴ = A² * A²  =  [ 0 -1 ]  *  [ 0 -1 ]  =    [ -1  0 ]

                          [ 1 0 ]      [ 1 0  ]        [  0  -1]

A⁶ =  A⁴ * A²  =  [ -1 0 ]  * [ 0 -1]  =  [ 0  1 ]

                          [ 0  -1 ]    [ 1 0 ]      [ -1 0]

A¹² = (A⁶)²  =   [ 0    1 ]

                           [ --1  0 ]

Note that   (A⁶)ⁿ  =  [ 0 1  ]          where n = 1,2,3,4.......

                                [ -1  0]

A¹⁰² = (A⁶)¹⁷  = [ 0 1 ]

                          [-1 0 ] 

B²  = [ 1 1  ]  *   [ 1 1  ]  = [ 1  2 ]

         [ 0   1 ]     [ 0  1]      [ 0  1]

B⁴  = [ 1 2 ]  * [ 1  2 ]  =   [1  4 ]

         [ 0  1]     [0  1 ]       [ 0  1]

B⁶  = B⁴ * B²  = [ 1 4 ]  *  [ 1 2 ]  =  [  1  6 ]

                         [ 0  1]      [ 0 1]       [   0  1]

Note that  B²ⁿ = [ 1  2n ]     where n =1,2,3,4.......

                          [ 0   1]

B¹⁰² = B^2*51  = [ 1  2*51 ]  =  [  1   102 ]

                           [ 0    1  ]       [  0     1  ]

A¹⁰²B¹⁰²   =  [  0  1 ]  *  [ 1   102 ]  =    [ 0       1 ]

                      [ -1  0 ]    [  0     1  ]         [  -1 -102 ]

B¹⁰²  A¹⁰² =  [ 1  102 ] *   [ 0 1  ]  =  [ -102   1 ] 

                     [ 0       1]      [ -1 0 ]      [  -1      0 ]

A¹⁰²B¹⁰² - B¹⁰²A¹⁰²  =  [  102     0  ]

                                       [   0    -102 ] 

See the following diagram

The slackrope walker is at "E"

We can form  two right triangles

One has a legs of (15 -3)   =  12   and 5

Then the hypotenuse  od this triangle =  sqrt [12^2 + 5^2] = sqrt (169) = 13  =  (DE)

This is part of the rope length

The other right triangle has legs of 12  and (14 - 5) = 9

The hypotenuse of this triangle is  sqrt [ 12^2 + 9^2]  =  sqrt [225]  = 15 = (BE)

This is the other part of the rope

The total length of the rope is (DE) + (BE) =    13 + 15  ≈  28 ft 

I think this  might be the question :

In the figure, angles ABC and AST are right angles. If AC = 35, AT = 11, and BT = 9, then what is AS?

AB = AT + BT =  11 + 9  = 20

Triangle AST  is similar to triangle ABC

AS / AT  =  AB / AC

AS / 11  = 20 / 35

AS = (11 * 20 ) / 35  =   220 / 35  =   44 / 7

I don't know anything about modular math, but here's an answer from a "Guest."

This one is relatively simple. You have to take each modulus x some constant + remainder = each other as follows:

(6 * a) + r =(20 * b) + 9 =(45 * c ) + 4=x. You simply try, by iteration, to find the values of a, b, and c.

If you look at (45*c) + 4 and try c= 1, then it becomes: 45*1+4=49. and if you look at 20b+9, you will see that b= 2, and it becomes 20*2 + 9 =49. Check and see if it balances 6*a +r. If you take 49 and divide by 6, you will get 8, because 6*8 =48. And so 49 mod 6 =1. Therefore, r can only =1.

Since the LCM of 6, 20, 45 =180, therefore x will be:

x =180C + 49, and C =0, 1, 2, 3, 4, 5.......etc. So:

x =180*0 + 49 =49

x =180*1 + 49 =229

x =180*2 +49 =409

x =180*3+49 =589..........and so on. All these values of x satisfy the congruences.

See proyaop's answer here : https://web2.0calc.com/questions/square_34

Sorry, I didn't know how to add one

https://web2.0calc.com/questions/algebra_10763

(n-3) + (n-2) + (n-1) + n + (n+1) + (n+2) + (n+ 3)  =  (28/9)(n + 3)

7n  = (28/9)(n + 3)

7n = (28/9)n + 28/3

63n = 28n + 84

63n - 28n = 84

35n = 84

n = 84 / 35  =  12 / 5

(n - 3)  = (12/5 - 3)  =  -3 / 5      (which is NOT an integer !!! )

Let A  =  [  1  0 ]             Let B  = [ a  b ]

              [  0  1 ]                           [ c  d ]

AB  =  [ a  b]

           [ c  d ]

BA  = [ a  b ]

          [ c  d ]