geno3141

In triangle(ABC), you have:  AB = 24,  AC = 28.  and  BC = 9.

Since you have the three sides, use the Law of Cosines to find angle(A)

  a²  =  b² + c² - 2·b·c·cos(A)

  9²  =  28² + 24² - 2·28·24·cos(A)

Solve this for angle(A).

Now, look at triangle(ADE).

AD = 35, AE = 30, and you have angle(A) from above.

Use the Law of Cosines for this triangle.

  a²  =  b² + c² - 2·b·c·cos(A)

  a²  =  35² + 30² - 2·35·30·cos(A)

Guest's answer is correct.

If you wish to do this problem algetraicly:

(n - 1)³  =  n³ - 3n² + 3n - 1

(n - 2)³  =  n³ - 6n²​ + 12n - 8

(n - 3)³  =  n³ - 9n²​ + 27n - 27

Therefore:

(n - 1)³ + (n - 2)³ + (n - 3)³  =  [ n³ - 3n² + 3n - 1 ] + [ n³ - 6n²​ + 12n - 8 ] + [ n³ - 9n²​ + 27n - 27 ]

                                            =  3n³ - 18n²​ + 42n - 36

Since n³  =  (n - 1)³ + (n - 2)³ + (n - 3)³,

          n³  =  3n³ - 18n²​ + 42n - 36

            0  =  2n³ - 18n²​ + 42n - 36

            0  =  n³ - 9n²​ + 21n - 18

Try various factors of -18 to find an integer divisor of  n³ - 9n²​ + 21n - 18.

Sooner, or later, you will get to  n = 6.

Since  n = 6  is a divisor,  n³ - 9n²​ + 21n - 18  =  (n - 6)(n² - 3n + 3)

Since  n² - 3n + 3  cannot be factored using only integers, the only answer is:  n = 6.

Do you mean to leave the value of BC to be blank?

Label the triangle as traingle(ABC), with A = left-bottom corner, B = right-bottom corner, and C = top corner.

Label the square as square(DEFG) with D = left-bottom corner, E = right-bottom corner, F on BC, and G on AC.

The bottom side of the triangle is now ADEB.

Let AD = 1.

Since angle(A) = 60^o and triangle(ADG) is a right triangle, AG = 2 and DG = sqrt(3).

Since DG = sqrt(3), DE, EF, and FG are all equal to sqrt(3).

This means that the square DEFG has an area of  sqrt(3) · sqrt(3)  =  3.

The bottom of the triangle, ADEB = AD + DE + EB.

AD = 1, DE = sqrt(3), and EB = 1, so ADEB = 2 + sqrt(3).

The formula for the area of an equilateral triangle is  A  =  sqrt(3) · s² / 3

--->   A  =  sqrt(3) · [ 2 + sqrt(3) ]² / 3  =  [ 7·sqrt(3) + 12 ] / 4

The ratio of the area of the square to the area of the triangle is:  3 : [ 7·sqrt(3) + 12 ] / 4

(a + 1/b)(b + 1/c)(c + 1/a)

=  [ (a + 1/b)(b + 1/c) ] (c + 1/a)

=  [ ab + a/c + 1 + 1/bc ] (c + 1/a)

=  abc + b + a + 1/c + c + 1/a + 1/b + 1/abc

=  abc + (a + b + c) + (1/a + 1/b + 1/c) + 1/abc

(1 + 1/a)(1 + 1/b)(1 + 1/c)

=  [ (1 + 1/a)(1 + 1/b) ]·(1 + 1/c)

=  [ 1 + 1/b + 1/a + 1/ab ]·(1 + 1/c)

=  1 + 1/c + 1/b + 1/bc + 1/a + 1/ac + 1/ab + 1/abc

=  1 + (1/a + 1/b + 1/c) + (1/ab + 1/ac + 1/bc) + 1/abc

Setting these two equal to each other:

abc + (a + b + c) + (1/a + 1/b + 1/c) + 1/abc  =  1 + (1/a + 1/b + 1/c) + (1/ab + 1/ac + 1/bc) + 1/abc

Subtracting like terms:

abc + (a + b + c)  =  1 + (1/ab + 1/ac + 1/bc)

Rewriting:

abc + (a + b + c)  =  1 + (c/abc + b/abc + a/abc)

Factoring:

abc + (a + b + c)  =  1 + (a + b + c)/abc

Since abc = 13:

13 + (a + b + c)  =  1 + (a + b + c)/13

Subtracting:

12 + (a + b + c)  =  (a + b + c)/13

Multiplying both sides by 13:

156 + 13(a + b + c)  =  (a + b + c)

Subtracting (a + b + c) from both sides:

156 + 12(a + b + c)  =  0

Subtracting 156 from both sides:

12(a + b + c)  =  -156

Dividing:

a + b + c  =  -13

You don't need a powerful calculator to get the answer that Guest gave you.

If you remember this from algebra;  a² - b²  =  (a + b)(a - b)

then          777777777777777² - 222222222222223² 

beomes:  ( 777777777777777 + 222222222222223 ) · ( 777777777777777 - 222222222222223 )

=                                        (  1 000 000 000 000 000 ) · ( 555 555 555 555 554 )

which gives you the digits from Guest's answer.

A system of two linear equations will have infinitely many solution if both equations reduce down to the same one.

3a + 4b  =  7

6a + 4b  =  k - 4b   --->   6a + 8b  =  k

Muliply the first equation by 2:  3a + 4b  =  7   --->   x 2   --->   6a + 8b  =  14

Compare that equation to the second equation:                       6a + 8b  =  k

They reduce to the same equation if  k = 14.

Because you have a sqrt(i) in the problem, let's start there ...

sqrt(i)  =  sqrt(2)/2 + (sqrt(2)/2)·i

Problem:  6 + 6·sqrt(3i)  =  6 + 6·sqrt(3)·sqrt(i)

  substituting:                      6 + 6·sqrt(3)·[ sqrt(2)/2 + (sqrt(2)/2)·i ]

  expanding:                        6 + 6·sqrt(6)/2 + 6·sqrt(6)/2 ·i

                                           ( 6 + 3·sqrt(6) ) + ( 3·sqrt(6) )·i

This is of the form  a + bi  where  a = ( 6 + 3·sqrt(6) )  and  b = ( 3·sqrt(6) )

We can now put this into  r·cis(theta)  form by using  r  =  sqrt( a² + b² )  and  theta  =  tan^-1( b/a )

r  =  sqrt[ ( 6 + 3·sqrt(6) )² + ( 3·sqrt(6) )²  =  sqrt( 144 + 36·sqrt(6) ]

theta  =  tan^-1[ ( 3·sqrt(6) ) / ( 6 + 3·sqrt(6) ) ]

Now, find the fourth root of r and divide theta by 4, to get your primary root; then, you can get the other three roots. 

3x² + 4x + 12  =  0

Use the quadratic formula to find the roots.

r  =  [ -4 + sqrt( 4² - 4·3·12 ) / ( 2·3 )  =  [ -4 + sqrt( -128 ) / ( 6 )   =  [ -4 + 8 · i · sqrt(2) ] / 6

s  =  [ -4 - sqrt( 4² - 4·3·12 ) / ( 2·3 )  =  [ -4 - sqrt( -128 ) / ( 6 )   =  [ -4 - 8 · i · sqrt(2) ] / 6

r²  =  ( [ -4 + 8 · i · sqrt(2) ] / 6 )²  =   [ -112 - 64·i·sqrt(2) ] / 36   (which can be reduced)

s²  =  ( [ -4 - 8 · i · sqrt(2) ] / 6 )²  =   ... 

r² + s²  =   ...