All Answers 356000 Answers

0.125 =(1/2)^(t/52500), solve for t

t =157,500 years

[32 + 1/43]  - [20 + 5/6  ]  =

[32 - 20 ]  +   [ 1/43 - 5/6 ]  =

[ 12 ] +  [ 6 - 43(5) ] / [ 43 * 6 ]  =

[12 ]  +  [ 6 -215 ] / 258  =

[12] +  [ -209 ]/ 258

[ 12 (258)  - 209 ] / 258  =

[3096  - 209 ] / 258  =

[2887 ]  /  258  =

11  + 49/258

The turtle 's change in position is    11  + 49/258   inches  deeper

One formula that you can use is:  A(t)  =  A₀·(½)^{( t / t_1/2 )}

where  t  =  time                                                       =  unknown

       A(t)  =  the amount remaining after time t         =  0.125

        A₀  =  the initial amount                                   =  1.000

       t_1/2  =  the half-life                                            =  52 500

A(t)  =  A₀·(½)^{( t / t_1/2 )}          --->     0.125  =  1.000 · (½)^{( t / 52 500 )}

                                            --->      0.125  =  (½)^{( t / 52 500 )}

                                      --->     log(0.125)  =  log( (½)^{( t / 52 500 )} )

                                     --->     log(0.125)  =  ( t / 52 500 ) · log( 1/2 )

                     --->      log(0.125) / log(1/2)  =  t / 52 50

    --->     [ log(0.125) / log(1/2) ] · 52 500  =  t

H=2; M=48; A=abs((30*H) - (5.5*M));print"A =", A,"Degrees")

A = 204 Degrees

The minute hand is at 48 minutes. That is clear. The hour hand has moved: 48 / 12 = 4  minutes after 2, which is 14 minutes from 12 O'clock. So, 48 - 14 =34 minutes, the angle between the two hands of the clock. But each minute on the clock is 360 /60 =6 degrees . Therefore, 34 minutes x 6 degrees =204 degrees.

The easiest way to do this is to actually draw the line  x = -3.

This line is a horizontal line that is three units below the x-axis.

Then drop each point as far below this line as the original point is above the line.

Point B, which is at (-6, -1), drops down to become the point (-6, -5).

Point A, which is at (-5, 2), drop down to become the point (-5, -8).

Point C is yours ...

Given:     v_i  =  11 m/s

               v_f  =  60 m/s

                a  =  8.0 m/s²

Using the formula:  (v_f)² - (v_i)²  =  2·a·d

               (60 m/s)² - (11 m/s)²  =  2·(8.0 m/s²)·d

          3600 m²/s² - 121 m²/s²  =  (16.0 m/s²)·d

                              3479 m²/s²  =  (16.0 m/s²)·d

       (3479 m²/s²) / (16.0 m/s²)  =  d

                                       217 m  =  d

I don't understand this:  "y=ab^xy=abx"  ...  do you mean  y  =  ab^x ?

If you do:  (0,9)     --->     9  =  a·b⁰

                             --->     9  =  a·1                       

                             --->     9  =  a     

                             --->     y  =  9·b^x 

       (3, 4608)   --->   4608  =   9·b³

                        --->     512  =  b³                            (dividing both sides by 9)

                        --->         8  =  b                              (taking the cube root of both sides)

I agree with CoolStuffYT that  n  >  8/3  and that  n  <  10.

Now, to finish you will need to count the positive integers between these two limits.

Thank you!

To understand the question better, it's easier to actually get an image, so I used : https://www.visnos.com/demos/clock

Lets solve for the hour. 

12 hours = 360°

1 hour or 60 mins = 30°

You would count all hours within the minute and hour hands. There are 5 hours in between. So because 1 hour is equal to 30°, you would multiply 5 and 30, which would then be 150°.

Let's solve for the minute.

This is where the image is good to have. It wants the obtuse angle, so you would subtract 60 from 48, which would leave you with 12 minutes left. Because 60 minutes is equal to 30°, you would then cross multiply with something that looked like this:

60  /  30      =      12  /  x

Solve for x and you would get 6°

Add your degrees 150° and 6°. Your answer is 156°. Voila! 

I'm sorry about that. I can program a computer to search for the answer, which is what I did. I'm not a mathematician, so I cannot give a mathematical solution. One of the mathematicians here may be able to solve it. Melody, CPhill, EP, Alan, heureka....etc. may take a look at it and see if they can solve it. Good luck.

Thx, Melody  !!!!

Thanks, CoolStuffYT  !!!!

If two pairs of jeans were  $94, then the cost per pair  =  94/2 =   $47

So....let  S  be  the cost of a shirt.....and we have that

4S  + 3(47)   =181

4S  + 141  = 181    

4S = 181 - 141

4S  =40  divide both sides  by  4

S = 10

The price  for a shirt was $10

That helps a lot. Thank you so much. :)

Lets see if I cant simplify this a bit and check it in the process.

There are 120 blocks

Let the first one be  MCZS  

M = material of the first one   \(n(M)= \frac{1}{2}\;\; of \;\;total, \qquad n(\bar M)=\frac{1}{2}\;\;of \;\;total \)

C = colour of the first one  \(n(C)= \frac{1}{3}\;\; of \;\;total, \qquad n(\bar C)=\frac{2}{3}\;\;of \;\;total \)

Z = size of the first one  \(n(Z)= \frac{1}{4}\;\; of \;\;total, \qquad n(\bar Z)=\frac{3}{4}\;\;of \;\;total \)

S= shape of the first one   \(n(S)= \frac{1}{5}\;\; of \;\;total, \qquad n(\bar S)=\frac{4}{5}\;\;of \;\;total \)

I will compare the second on to this one,

There are 6 ways that exactly 2 properties can be the same.

\(n(MC\bar Z \bar S)=120*\frac{1}{2}*\frac{1}{3}*\frac{3}{4}*\frac{4}{5}=12\\~\\ n(M\bar C Z \bar S)=120*\frac{1}{2}*\frac{2}{3}*\frac{1}{4}*\frac{4}{5}=8\\~\\ n(M\bar C \bar Z S)=120*\frac{1}{2}*\frac{2}{3}*\frac{3}{4}*\frac{1}{5}=6\\~\\ n(\bar M C Z \bar S)=120*\frac{1}{2}*\frac{1}{3}*\frac{1}{4}*\frac{4}{5}=4\\~\\ n(\bar MC\bar Z S)=120*\frac{1}{2}*\frac{1}{3}*\frac{3}{4}*\frac{1}{5}=3\\~\\ n(\bar M \bar C Z S)=120*\frac{1}{2}*\frac{2}{3}*\frac{1}{4}*\frac{1}{5}=2\\~\\\)

\(12+8+6+4+3+2=35\)

After the first block was taken there was 119 mot blocks to choose from so 

The probability that the 2 blocks have exactly 2 properties in common is         \(\boxed{\frac{35}{119}}\)

Now our answers agree.    

Coding:

n(MC\bar Z \bar S)=120*\frac{1}{2}*\frac{1}{3}*\frac{3}{4}*\frac{4}{5}=12\\~\\

n(M\bar C Z \bar S)=120*\frac{1}{2}*\frac{2}{3}*\frac{1}{4}*\frac{4}{5}=8\\~\\

n(M\bar C \bar Z S)=120*\frac{1}{2}*\frac{2}{3}*\frac{3}{4}*\frac{1}{5}=6\\~\\

n(\bar M C Z \bar S)=120*\frac{1}{2}*\frac{1}{3}*\frac{1}{4}*\frac{4}{5}=4\\~\\

n(\bar MC\bar Z S)=120*\frac{1}{2}*\frac{1}{3}*\frac{3}{4}*\frac{1}{5}=3\\~\\

n(\bar M \bar C Z S)=120*\frac{1}{2}*\frac{2}{3}*\frac{1}{4}*\frac{1}{5}=2\\~\\

I just wrote out g(x)'s polynomial form. In other words, because I didn't know the actual coefficients, I substituted with variables like a₀, a₁, etc. to denote the coefficient for that x term

Your friend has an egg collection comprising at least 200 eggs.
He wants to store them in dozen-egg containers.
After filling as many containers as possible, the last container had 1 egg left over.
He then decided to store his eggs in customized baker-dozen-egg containers, where each container can hold 13 eggs.
It turns out that, after filling as many of these containers as possible, he still has 1 egg left over.
What is the minimum number of eggs that your friend could have?

\(\begin{array}{|rcll|} \hline \text{eggs}_{\text{min.}} &=& 1+lcm(12,13)*n \qquad n\in \mathbb{N} \\ \text{eggs}_{\text{min.}} &=& 1+12*13*n \\ \text{eggs}_{\text{min.}} &=& 1+156*n \quad | \quad \text{eggs}_{\text{min.}} \mathbf{\geq 200 !},\ \text{so } n=2 \\ \text{eggs}_{\text{min.}} &=& 1+156* 2 \\ \mathbf{\text{eggs}_{\text{min.}}} &=& \mathbf{313} \\ \hline \end{array}\)