CPhill

Working on them  one at a time, RP....give me a few....

2. An ammonia and water mixture fills a 5 gallon container. 80% of the mixture is ammonia but some of the mixture will be drained and replaced with pure water. If a 5 gallon mixture of fifty percent ammonia is desired, how many quarts of the mixture need to be drained before the water is added? (A quart is one-fourth of a gallon.) Express your answer as a decimal to the nearest tenth.

5 gallons  = 20 qts  ....and 80% ammonia   = 20% water

Let x be the number of quarts of mixture that we need to drain

So...when this is done we have    [20 -x] qts of 20% water

[20 (.20) - x(.20)]  =

[20 - x] .20

To this, we obviously need to add back  "x" quarts of  pure water  to get 20 qts of  50% water [ = 50% ammonia ]

So we have

[20 - x ] .20  +  x (1)  = 20(.50)  simplify

4  - .2x + x  = 10

.8x + 4  = 10    subtract  4 from each side

.8x + 6     dibide both sides by .8

6 / [8/10]  = x

60/ 8  = x

x  = 7.5

So...we need to drain 7.5 qts  of the 20%water soluiton...and then add back this many qts of pure water to get a 20 qt solution of 50% water

1. We bought 2 pens and 6 pencils, the ratio of the amount we spent on pencils to the amount we spent on pens was 7:4. Later we went back and bought 4 pencils and 3 pens and spent 1 dollar less. How much money did we spend that time?

Let the cost of a pen  = x  and the cost of  a pencil  = y

So we have that  

6y / 2x  = 7 / 4

3y / x  = 7/4

3y  = (7/4)x

y = (7/12)x

And we know that

2x + 6y   = A        where A  is in dollars

2x + 6(7/12)x  = A

2x + (42/12)x  = A

2x + 3.5x  = A

5.5x   = A  = (11/2)x    (1)

And we also know that

3x + 4y  =  A  - 1

3x + 4(7/12)x  = A  - 1

3x + (28/12)x  = A  - 1

3x + (7/3)x   = A  - 1

(16/3)x  = A  - 1      sub( 1)  in for  A

(16/3)x = (11/2)x- 1    subtract (11/2)x from both sides

-1/6x  = -1

x  = 6 (dollars)  = cost of a pen

And y = (7/12)(6)  = 42/12 =  3.5  dollars = cost of a pencil

So...the first time we spent

2(6) + 6(3.5) = $33

And the second time we spent

3(6) + 4(3.5)  = $32

Let x  be the number  of software programs produced  and y be the number of video games produced

And we have these  inequalities

x <=250

y <=325

And 

x + y < = 450

And we wish to maximinze this objective proift function, P :

P  = 65x  + 40y

Here's  the graph : https://www.desmos.com/calculator/k7ivq8jnv2

The max profit will occur at the corner point  ( 250, 200)

So...produce 250 software programs and 200 video games 

Let  PQ  be one leg and QR  be the other leg

PQ  = sqrt  [ 6^2 + ( 9- 5)^2 ] =   sqrt [ 36  + 4^2 ] = sqrt [52]

QR  = sqrt [ (12 - 6)^2  + 9^2  ]  =  sqrt  [ 6^2 + 9^2 ] =  sqrt [ 117]

And the area of a right triangle  = (1/2) (product of the leg lengths)  =

(1/2)sqrt(52)sqrt(117)  = (1/2) sqrt(6084)  = (1/2) 78   =   39 units^2

(3)/(2)-(10)/(3)((13)/(2)n-(1)/(3))>=-(1513)/(18)

I'm gonig to assume  that this is :

(3/2) - (10/3)[ (13/2)n - (1/3) >=  - (1513/18)  distribute the (10/3)  over the terms in parentheses

(3/2) - (130/6)n  +  10/9  >=  - (1513 / 18)      multiply through by LCM  of  2, 6, 9 and 18  = 18

27  - 390n + 20  >= - 1513   combine like terms

47 - 390n  > = -1513       subtract 47 from  both sides

-390n > = -1560    divide both sides by -390  and reverse the ineqaulity sign

n <= 4

We have that

x + y  =  5.8  ⇒    y  = 5.8 - x    (1)

And note that rate  * time  = distance covered.....so..

5.5x + 13.5y  = 40    (2)      put (1)  into (2)  for  y   and we have

5.5x + 13.5(5.8 - x)   = 40    simplify

5.5x  + 78.3  - 13.5 x   =  40

-8x + 78.3  = 40       subtract  78.3 from both sides

-8x  = - 38.3      divide both sides by  -8

x ≈  4.8  hrs  running

And  y ≈  5.8 - x  =  5.8 - 4.8  ≈  1 hrs biking

Notice that, at 7.5 miles....that the cost of taxi "A"  becomes cheaper than the cost of taxi "B"

So...taxi "B"  is the better deal for the first 7 miles ⇒  "D"

15.  -3x ( 5x + 1) (8 - 2x)  = 

-3x ( 5x * 8 + 1*8 - 2x*5x - 2x*1) =

-3x ( 40x + 8  - 10x^2  - 2x)  =

-120x^2  - 24x  + 30x^3  + 6x^2  =

30x^3 - 114x^2  - 24x 

"A" will  produce  vertical "stretching " by a factor  of 4