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AC = 24.

I can do this!

1) In how many ways can I distribute 6 identical gummy bears and 6 identical chocolates to 4 students, if each student must receive at least 1 of each type of candy?

By stars and bars, the number of ways is c(8,3)*c(8,3) = 3136.

2) In how many ways can I distribute 6 identical gummy bears and 6 identical chocolates to 4 students, if each student must receive exactly 3 pieces of candy?

By stars and bars, this is c(14,4) = 1001.

Arctan 2/3 = atan(0.6666666666666667) = 33.69006752598 degrees

v = sqrt (3^2 + 2^2) = sqrt 13     

       sqrt13 cos 33.69  , sqrt 13 sin 33.69

would also like to know this

Set the y equations equal

24+ 1/8 t = 18+ 1/4 t      solve to find t = 48    (4.8 seconds)

Sub this value into the x equations ......if the results are equal a collision happens at  4. 8 sec

AB and AC are common tangents from point A to a circle, where B and C lie on the circle.  P is on AB, and R is on AC so that PR is tangent to the circle.  The perimeter of triangle APR is 30, and BC = 10.  Find the perimeter of triangle ABC.

Hint: 

If AP = AR, then PC = 9.357483637

Area of Δ APR = 40.98780306 u²  

Solve:  |x - 1| / |x + 3|  <  1

Multiply both sides by |x + 3|   --->   |x - 1|  <  1 · |x + 3|   --->   |x - 1|  <  |x + 3| 

There are up to four possibilities:

1)  x - 1 <= 0   --->   x <+ 1     and     x + 3 <= 0   --->   x <= -3      --->     x <= -3

     --->   -(x - 1) < -(x + 3)

              -x + 1  <  -x - 3

                     1  <  -3               

2)  x - 1 <= 0   --->   x <= 1     and     x + 3 >= 0   --->   x >= -3     --->     -3 <= x <= 1

      --->   -(x - 1)  <  x + 3

                -x + 1  <  x + 3

                     -2  <  2x

                     -1  <  x            --->     x > -1

3)  x - 1 >= 0   --->   x >= 1     and     x + 3 <= 0   --->   x <= -3     

4)   x - 1 >= 0   --->   x >= 1     and     x + 3 >= 0   --->   x >= -3     --->     x >= 1

     --->     x - 1 <  x + 3

                   -1  <  3                 1>

Combining  x > -1  or  x >1   --->   x > -1

B=(7,4) and C=(3,7)

The change of base formula:  log_bx  =  log_nx / log_nb.

Example:  log₂8  =  log8 / log2

                 log₂8  =  ln8 / ln2

Based on the appearance of the "link" you sent, this is probably a file buried deep inside of your computer. Of course, I do not have access to that, so I (nor can anyone else) can help you. Maybe try uploading the image here.

x  =  t + 3   --->   t  =  x - 3

y  =  2t² - 4   --->   y  =  2(x - 3)² - 4   --->   y  =  2(x² - 6x + 9) - 4   --->   y  =  2x² - 12x + 14

Since the minimum t-value is -4 and x = t + 3, the minimum x value will be -4 + 3  =  -1.

Since the maximum t-value is 0 and x = t + 3, the maximum x-value will be 0 + 3  =  3.

The vector from (a,b) to (c,d) is (c - a, d - b).

--->   (1 - -3, -6 - 2)  =  (4, -8)

If set P has 10 numbers with an arithmetic mean of p, the sum of those 10 numbers is 10p.

If set Q has 20 numbers with an arithmetic mean of q, the sum of those 20 numbers is 20q.

To find the arithmetic mean of all thirty numbers:  (10p + 20q) / 30/

Rectangles ABCD and ANME are similar.

AE is 1/2 of AD

AG must be 1/2 of AE    

thank you for always helping me!! i appreciate it lots.

https://web2.0calc.com/questions/i-really-need-help-im-so-confused

The point that minimizes PR + RQ is the point on the line PRQ.

First, let's find the equation of the line:  y - -3  =  (1 - -3)/(4 - -2)·(x - -2)

                                                               y + 3  =  (4/6)·(x + 2)

                                                               y + 3  =  (2/3)·(x + 2)

For the point (0,k), x = 0:                        y + 3  =  (2/3)·(0 + 2)

                                                               y + 3  =  4/3

                                                                     y  =  -5/3      --->   k  =  -5/3

No, that answer's wrong. I know because I found the correct answer by myself! :D I guess I just needed a little more time.

The real answer is 4/25 if anyone's wondering.

The largest integer appears to be 8. It divides all integers of the form:7^(2n + 1) + 1 from n=1 to infinity.

To find the height of the kite, use the sin:  sin(x)  =  opposite / hypotenuse

   --->   sin(27)  =  height / 176   --->   176 · sin(27)  =  height

To find the horizontal distance, use the cos:  cos(x)  =  adjacent / hypotenuse

   --->   cos(27)  =  horizontal / 176   --->   176 · cos(27)  =  horizontal

y = 3/4 * root((3*x^4)/c^2,3)

c is the speed of light in a vacuum.  If X and Y are distances measured in meters, does this equation make sense?

Hello Guest!

Is your equation called that

\(y=\frac{3}{4\cdot \sqrt{\frac{3x^4}{c^{2.3}}}}\)    (according to rule point calculation operates from left to right. 2,3 means 2.3?)

or so

\(y=\frac{3}{4}\cdot \sqrt{\frac{3x^4}{c^{2.3}}}\)  ?

I assume the second equation.

\(\color{BrickRed}y=\frac{3}{4}\cdot \sqrt{\frac{3x^4}{c^{2.3}} }\)

\(y=\frac{3}{4}\cdot \sqrt{\frac{3x^4}{c^{2.3}} }\\ y^2=\frac{9}{16}\cdot \frac{3x^4}{c^{2.3}}\\ c^{2.3}=\frac{9}{16}\cdot \frac{3x^4}{y^2}\)

\(c=(\frac{9}{16}\cdot \frac{3x^4}{y^2})^{-2.3}\)

The equation doesn't make sense.
The unit of the speed of light is \(m/sec\ (300 \times 10 ^ 9\ m / sec).\)

The unit on the right is \(m^{-4.6}\), no matter what is used for x and y.

\(c\color{red}\neq\)\((\frac{9}{16}\cdot \frac{3x^4}{y^2})^{-2.3}\)

\(y\color{red}\neq\)\(\frac{3}{4}\cdot \sqrt{\frac{3x^4}{c^{2.3}}}\)

  !

j         k

12     4
6       6
4      12

Add them all up