All Answers 358483 Answers

Drop back 10 yards and punt.........!!!!

OK, seriously  ......ask someone here.......

Correct! Can you solve Riddle 2?

8)  This is like 7

.10 ( 16 - x)  + 1x  =  .35(16)   simplify

1.6 - .10x  + x  =  5.6      

1.6  + .90x   = 5.6      subtract 1,6 from both sides

.90x  = 4    divide both sides by .90

x = 4/.90   =  [40/9] quarts need to be drawn off   ≈  4.44 quarts

The length of the longest side cannot be as long as, or longer than the two shorter sides added together.

Another way to say this is to say that the lonest side cannot be half, or more than, the total perimeter.

So, the longest side cannot be 0.7 km, but can be any positive fraction smaller than this.

The other two must total to the remaining amount when added together.

So, if you choose the longest side to be 0.6 km, the other two must total to 0.8 km (1.4 - 0.6 = 0.8).

One of the other two sides could be 0.5 while the other could be 0.3 (neither one of these can be larger than 0.6 because, for this example, the longest side is assumed to be 0.6).

There is an infinite number of possibilites which satisfy these restrictions.

7)  80% cleansing powder = 20% water  = .20      60% cleansing powder = 40% water  = .40      pure water = 100%  = 1

This one is a little trickier......I hope I've done it correctly....!!!

Let's call the amount we draw off = x       And this is = to the amount of  pure water we replace it with....so we have....

.20water  * [ starting amount - amt drawn off]  plus   "x" amt of pure water =  40% water * 200L
 

.20(200-x) + 1x  =  .40 [ 200]    simplify

40 - .20x  + 1x   = 80   

40 + .80x  = 80       subtract 40 from each side

.80x  = 40            divide both sides by ..80

x = 40/.80   = 50 mL   = amt drawn off

.

Simplify the following:
16/4+5×6-(3+4+1)

The gcd of 16 and 4 is 4, so 16/4 = (4×4)/(4×1) = 4/4×4 = 4:
4+5×6-(3+4+1)

3+4+1 = 8:
4+5×6-8

5×6  =  30:
4+30-8

4+30 = 34:
34-8

| 2 | 14
| 3 | 4
- |  | 8
| 2 | 6:
Answer: | =26
 

Riddle 1:

Assuming that Guard A is guarding Door A and Guard B is guarding Door B.

Go up to guard A and ask "Will Guard B tell me that you (Guard A) are guarding the door to eternal bliss?".

If Guard A always tells the truth and Door A is the door to eternal bliss, then Guard A will answer "No" because Guard B will lie.

If Guard A always lies and Door A is the door to eternal bliss, the Guard A will answer "No" because Guard A lies.

If Guard A always tells the truth and Door A leads to eternal punishment, then Guard A will answer "Yes".

If Guard A always lies and Door A leads to eternal punishment, the Guard A will answer "Yes".

So, if the answer is "No", choose that door; if the answer is "Yes", choose the other door.

(It doesn't make any difference which Guard and which Door you approach.)

In PTA fundraising activity, couples who came were charged $400. However if only one parent came, he was charged $250.00.

59 People attended the activity and $12,150 was raised. How many couples attended?

Let C=number of couples

Let S=number of singles

2C + S=59,and,

400C + 250S=12,150

Solve the following system:
{2 C+S = 59 |     (equation 1)
400 C+250 S = 12150 |     (equation 2)
Swap equation 1 with equation 2:
{400 C+250 S = 12150 |     (equation 1)
2 C+S = 59 |     (equation 2)
Subtract 1/200 × (equation 1) from equation 2:
{400 C+250 S = 12150 |     (equation 1)
0 C-S/4 = (-7)/4 |     (equation 2)
Divide equation 1 by 50:
{8 C+5 S = 243 |     (equation 1)
0 C-S/4 = -7/4 |     (equation 2)
Multiply equation 2 by -4:
{8 C+5 S = 243 |     (equation 1)
0 C+S = 7 |     (equation 2)
Subtract 5 × (equation 2) from equation 1:
{8 C+0 S = 208 |     (equation 1)
0 C+S = 7 |     (equation 2)
Divide equation 1 by 8:
{C+0 S = 26 |     (equation 1)
0 C+S = 7 |     (equation 2)
Collect results:
Answer: |  C=26      and      S=7
 

There can be two equations with two variables:  C = number of couples; S = number of singles.

One equation is for the number of people:  2C + S  = 59

(59 is the total number of people, use '2C' because each couple contains two persons.)

The other equation holds the amount of money made:  400C + 250S  =  12,150

Combining these two equations:                                    

Keep the money equations                             :   400C + 250S  =  12,150

Mutiply the number of people equation by 200:  400C + 200S  =  11,800

Subtract down the columns:                                              50S  =  350

Divide by 50:                                                                         S  =  7  (there were 7 single parents)

Since  2C + S  =  59   --->   2C + 7  =  59   --->   C  =  26                  (there were 26 couples)

6)   I'm assuming that we want to end up with a solution that is 75% salt.....this is like 4.....and 75% salt = 25% water....so we have

50% solution of water* 4 L   -  100% water * unknown amt evaporated   =  25% water [ 4L  - unknown amt]

.50(4)   -  1x  = .25 [ 4 - x]     simplify

2 - 1x  = 1  - .25x       add 1x  to both sides,  subtract 1 from both sides

1  = ..75x   = (3/4)x       multiply both sides by  4/3

[4/3 ] L  = x      this is the amt to be evaporated

.

.

A formula for exponential growth and decay is:  A  =  A₀e^kt

where A is the final amount, A₀ is the initial amount, k is the proportionality constant, and t is the time.

For half life, A = 1/2A₀.

So, substituting into the original formula:  1/2A₀ = A₀e^kt

Dividing both sides by A₀:              1/2 = e^kt

Since the time is 1600 years:        1/2 = e^1600k

Take the ln of both sides:         ln(1/2) = ln(e^1600k)

Inside an ln expression, the exponent comes out as a multiplier:   ln(1/2)  =  1600k ln(e)

ln(e) = 1                                                                                          ln(1/2)  = 1600k

Divide both sides by 1600 and simplify:     k = -0.000433

The formula becomes:  A = A₀e^-0.000433t

Let's assume we start with an initial amount of 1 (100%)  --->   A = 1e^-0.000433t

After 2.07 x 10³ years                                                        --->   A  =  e^{-0.00043(2070)}

So, the amount left is .408, or about 41% of the original amount.

Mister Adam sells tapioca pearls in 4 kilo and 3 kilo containers. With much demand for the tapioca pearls, he sold 30 containers weighing 100 kilos. How many of each type did he sell?

Let f=4 kilo containers

Let t=3 kilo containers

f + t=30, and,

4f + 3t=100

Solve the following system:
{f+t = 30 |     (equation 1)
4 f+3 t = 100 |     (equation 2)
Swap equation 1 with equation 2:
{4 f+3 t = 100 |     (equation 1)
f+t = 30 |     (equation 2)
Subtract 1/4 × (equation 1) from equation 2:
{4 f+3 t = 100 |     (equation 1)
0 f+t/4 = 5 |     (equation 2)
Multiply equation 2 by 4:
{4 f+3 t = 100 |     (equation 1)
0 f+t = 20 |     (equation 2)
Subtract 3 × (equation 2) from equation 1:
{4 f+0 t = 40 |     (equation 1)
0 f+t = 20 |     (equation 2)
Divide equation 1 by 4:
{f+0 t = 10 |     (equation 1)
0 f+t = 20 |     (equation 2)
Collect results:
Answer: | f=10    and    t=20
 

Thank you very much.

the model???

like it has to equal like this:

P =

this is the problem:

A school board has a plan to increase participation in the PTA. Currently only about 29 parents attend meetings. Suppose the school board plan results in logistic growth of attendance. The school board believes their plan can eventually lead to an attendance level of 87 parents. In the absence of limiting factors the school board believes its plan can increase participation by 10% each month. Let m denote the number of months since the participation plan was put in place, and let P be the number of parents attending PTA meetings.

What is the carrying capacity K for a logistic model of P versus m?

K=87

(b) Find the constant b for a logistic model.

b=2

(c) Find the r value for a logistic model. Round your answer to three decimal places.

r=.095

QUESTION:

(d) Find a logistic model for P versus m.

Solve for m:
3.3489 m = 0.13175/m^0.575

3.3489 m = (33489 m)/10000 and 0.13175/m^0.575 = 527/(4000 m^(23/40)):
(33489 m)/10000 = 527/(4000 m^(23/40))

Cross multiply:
133956000 m^(63/40) = 5270000

Divide both sides by 133956000:
m^(63/40) =0.03934127........

m=0.03934127^(40/63)=0.12818590744853........

5)  All of these work the same, pretty much....

% of 25% protein milk * unkown amt  + % of 40% protein milk * [ 12 - unknown amt ]   =  40% protein milk * 12 L

.25x  + .40(12 - x) = .35(12)    simplify

.25x + 4.8 - .40x =  4.2         

-.15x + 4.8 = 4.2      subtract  4.8 from both sides

-.15x  = -0.6      divide both sides by  -.15

x = 4  L of 25%    and    12 - 4   = 8L  of 40%

4).....How much water = x  .....it's easier if we express everything in terms of % water....... note that 32% salt solution = 68% water solution    and pure water = 100%  = 1     and a 50% solution = .50 water

.68(20)  -  1x  =  .50 [ 20 - x]    simplify

13.6 - 1x  = 10 - .50x      subtract 10 from both sides.....add 1x  to both sides

3.6 =  .50x       divide both sides by .50

x  = 3.6 / .50   = 7.2 L

But I just tried Melody's answers ...and they are not equal at the end of th Five year period?

Look at her 3% value     at the end of five years it will be 22811.55 (1.015)^10 = 26473.73

then look at the 7% one                                                    16660.32 (1.035)^10 =  23502.43      They are not equal?   Am I mireading the question?

I thought the neding values were supposed to be equal.......Did I miss calculate?   If you do these calcs with my numbers, the ending values are all equal.....What did I do wrong?  AM I wrong????

~jc

3) Note.... 1/2 = .50       3/4 = .75     3/5 = .60   ....so we have

%purity * known amount of XRD  + %purity * unknown amount of QXJ  =  final %purity [ known + unknown amts. ]

.50(1)  + .75x  = .60 (1 + x)      simplify

.50  + .75x  = .60 + .60x        subtract .60x, 50   from both sides

.15x  =   .10      divide both sides by .15

x = [.10 / .15 ] L  =  [10/15 ] L =  [2/3]L

1.) I understand what you're saying and I go the correct answer. :)

2.) I'm sorry for another typo. The question was supposed to be:

The length of a picture is 4/3 times its width. It is placed in a picture frame of uniform width of 2 cms. If the are of the frame is 100 sq.cms., what are the dimensions of the picture?

Sorry for the typo. It's 81 sq.cms.. 

Thank you for the help.

What part are you having problems with?   

I think you are just beginning....wait until you get to Fourier Transforms or Advanced Applied Mathemeatics !  (or Special Relativity, Electromagnetics etc)      Yikes. 

    Cool emoji !

2)   Note that pure octane = 100% = 1.00   

.95(1)   + 1.00 (x)   = .96(1 + x)   simplify

.95  + x   =  .96 +  .96x      subtract  .96x,  .95 from both sides

.04x  = .01   divide both sides by  .04

x = 1/4 L

thats wrong ^ i keep getting it wrong..