geno3141

Let  c  be the cost of renting 1 chair.

Let  t  be the cost of renting 1 table.

     2c + 5t  =  45.00   --->  multiply by 4     --->   8c + 20t  =  180.00

     8c + 3t  =  44.00   --->   multiply by -1   --->  -8c  -   3t  =  -44.00

Add down the columns:                                               17t  =  136.00

Divide by 17:                                                                   t  =  8.00

Substitute this value into  2c + 5t  =  45.00   --->   2c + 5(8.00)  =  45.00

                                                                                 2c + 40.00  =  45.00

                                                                                              2c  =  5.00

                                                                                                c  =  2.50

An appropriate formula is:  A  =  P·(1 + r)ⁿ

where  A  =  final amount                       =  $15,000.00

            P  =  principal  =  initial amount  =  $7500.00

            r  =  yearly rate, as a decimal     =  0.07

            n  =  number of years                 = 

--->                  15,000  =  7500(1 + 0.07)ⁿ 

--->                           2  =  1.07ⁿ                      (divide both sides by 7500)

--->                    log(2)  =  log(1.07ⁿ)               (because the unknown is in the numerator, take the log of

                                                                               both sides)

--->                    log(2)  =  n·log(1.07)              (property of logs)

--->   log(2) / log(1.07)  =  n                             (divide both sides by log(1.07))

If you finish this, you will have your answer.

Let  t  be the ten's digit of the original number and  u  be the unit's digit of the original number.

This means that the original number is  10·t + u.

Reversing the digits, the new number is  10·u + t

Adding these numbers together, we get a sum of 77:  (10t + u) + (10u + t)  =  77

Simplifying:                                                                                    11t + 11u  =  77

Dividiing each term by 11:                                                                     t + u  =  7

Subtracting these numbers, we get a difference of 45:  (10t + u) - (10u + t)  =  45

Simplifying:                                                                                         9t - 9u  =  45

Dividing each term by 9:                                                                         t - u  =  5

Combining these two equations:            t + u  =  7

                                                                t - u  =  5

Adding down the columns:                         2t  =  12

Divide by 2:                                                  t  =  6

Substituting back into the equation  t + u  =  7   --->   6 + u  =  7   --->   u  =  1

You'll need to write the two numbers ...

Let           x  =  pounds of the coffee worth $1.92 per pound

then  30 - x  =  pounds of the coffee worth $2.72 per pound

   cost x amount  +  cost x amount  =  final cost x final amount

                  1.92·x  +  2.72·(30 - x)  =  2.24·(30)

                    1.92x + 81.60 - 2.72x  =  67.20

                                 81.60 - 0.80x  =  67.20

                                            -0.80x  =  -14.40

                                                    x  =  18 pounds of coffee worth $1.92 per pound

Now, you will need to find the number of poiunds of he coffee worth $2.72 per pound.

Amount earned  =  hours x dollars per hour.

You:        (t - 6)(2t - 5)  =  2t² - 17t + 30

Andrew:  (2t - 8)(t - 5)  =  2t² - 18t + 40

Since the two of you earned the same amount, set these equations equal to each other:

                                            2t² - 17t + 30  =  2t² - 18t + 40

Subtract 2t² from both sides:     -17t + 30  =  -18t + 40

and, continue solving ...

f(x)  =  (-4x² -4x + 48) / (2x² - 10x + 12)

a)  Divide both the numerator and denominator by x²   --->   (-4 - 4/x + 48/x²) / (2 - 10/x + 12/x²)

     as x approaches infinity, every term with an x in the denominator goes to zero; so --->  -4/2  =  -2

b)  Horizontal asymptote:  y  =  -2

c)  y-intercept:  x = 0   --->   y = 4

d)  Factor the numerator:  -4x² - 4x + 48  =  -4(x² + x - 12)  =  -4(x + 4)(x - 3) 

     Factor the denominator:  2x² - 10x + 12  =  2(x² - 5x + 6)  =  2(x - 2)(x - 3) 

     Since both the numerator and denominator has a factor of 3, a removable discontinutity occurs at  x= 3

     A vertical asymptote occurs at x = 2.

e)  x-intercept:  y = 0   --->   x = -4

              a^b  =  b^a 

      log(a^b)  =  log(b^a)

     b·log(a)  =  a·log(b)

Since  b = 9a:

       9a·log(a)  =  a·log(9a)

Divide both sides by a:

         9·log(a)  =  log(9a)

           log(a⁹)  =  log(9a)

                  a⁹  =  9a

Either a = 1  (which won't work)

or:         a⁸  =  9

               a  =  9^1/8

               a  =  (3²)^1/8

                a  =  3^1/4

and  b  =  9·3^1/4

Which means that I can't get the answer that you want!

Let's find the vertex by completing the square.

First, move the 18 to the other side:     3y - 18  =  2x² - 16x

Factor out the 2:                                   3y - 18  =  2(x² - 8x)

Complete the square:                           3y - 18  =  2(x² - 8x + 16) - 32

Move the -18                                               3y  =  2(x - 4) - 14

Divide both sides by 3:                                  y  =  (2/3)(x - 4)² - 14/3

The vertex is  (4, -14/3)

2log(a - 2b)  =  log(a) + log(b)

Using properties of logs:

--->          log(a - 2b)²  =  log(a·b)

--->               (a - 2b)²  =  ab

--->      a² - 4ab + 4b²  =  ab

--->      a² - 5ab + 4b²  =  0

--->        (a - 4b)(a - b)  =  0

Either  a - 4b  =  0   --->   a = 4b   --->   a/b  =  4

or          a - b  =  0   --->   a  =  b     [but this is impossible for it would make a - 2b negative]

Each of these logs can be written into log form (that is, log₁₀ form).

(log₂x)(log₃x)(log₄x)(log₅x)  =  [log(x)/log(2)]·[log(x)/log(3)]·[log(x)/log(4)]·[log(x)/log(5)]

                                            =  [log(x)]⁴ / [log(2)·log(3)·log(4)·log(5)] 

(log₂x)(log₃x)(log₄x)  =  [log(x)/log(2)]· [log(x)/log(3)]· [log(x)/log(4)]

                                  =  [log(x)]³ / [log(2)·log(3)·log(4)]

] 

(log₂x)(log₃x)(log₅x)  =  [log(x)/log(2)]· [log(x)/log(3)]· [log(x)/log(5)]

                                  =  [log(x)]³ / [log(2)·log(3)·log(5)]

(log₂x)(log₄x)(log₅x)  =  [log(x)/log(2)]· [log(x)/log(4)]· [log(x)/log(4)]

                                  =  [log(x)]³ / [log(2)·log(4)·log(5)]

(log₃x)(log₄x)(log₅x)  =  [log(x)/log(3)]· [log(x)/log(4)]· [log(x)/log(4)]

                                  =  [log(x)]³ / [log(3)·log(4)·log(5)]

To get rid of denominators, multiply both sides by [log(2)·log(3)·log(4)·log(5):

--->   [log(x)]⁴  =  [log(x)]³ · [log(5)]  +  [log(x)]³ ·[log​(4)]  +  [log(x)]³ ·[log​(3)]  +  [log(x)]³ ·[log​(2)]

                       =  [log(x)]³ · [ log(5) + log(4) + log(3) + log(2) ]

Divide both sides by [log(x)]³ 

--->   log(x)  =  log(5) + log(4) + log(3) + log(2)

                    =  log( 5 · 4 · 3 · 2 )

                    =  log( 120 )

--->  x = 120