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Sum A =5050
Sum B =15050
GCD(5050, 15050) = 50

Let's make the solution super-formal!  We can write -4 in exponential notation as 4e^(pi*i), so the equation is z^4 = 4e^(pi*i).

By Hamilton's Theorem, the solutions are z = 4^{1/4}*e^(pi*i/4), 4^{1/4}*e^(pi*i/4 + pi/4), 4^{1/4}*e^(pi*i/4 + 2*pi/4), and 4^{1/4}*e^(pi*i/4 + 3*pi/4).  Since 4^{1/4} = sqrt(2) and e^(pi*i/4) = (1 + i)/sqrt(2), the first solution is 1 + i.  Then the other roots work out as

4^{1/4}*e^(pi*i/4 + pi/4) = 1 - i,

4^{1/4}*e^(pi*i/4 + 2*pi/4) = -1 - i, and

4^{1/4}*e^(pi*i/4 + 3*pi/4) = -1 + i.

The remainder is 3x^2 + 4x - 1.

work backwards from 2020

remember if you go backwards then you can -1 and divide by 5

It falls out easily.

Once you can do it backwards you can then reverse everything and do it forwards.

Please no one provide a fuller answer. Guest needs to do some thinking on his/her own.

I also think you can use stars and bars method but I think the answer is  54C4 = 316251

Let x = tan(arcsin(0.2)).  Compute x^2.

Let 

\(\theta=asin(\frac{1}{5})\\ \theta \text{ is in the 1st or 2nd quadrant}\\ \text{Draw a right angled triangle for the first quadrant version}\\ \text{opp=1, hypotenuse=5, adj=} \sqrt{24}\\ x=tan\theta =\pm \frac{1}{\sqrt{24}}\\ x^2=\frac{1}{24} \)

LaTex:

 quadrant version}\\ \text{opp=1, hypotenuse=5, adj=} \sqrt{24}\\ x=tan\theta =\pm \frac{1}{\sqrt{24}}\\ x^2=\frac{1}{24} 

See the answer here: https://web2.0calc.com/questions/help-please_92732#r4 

By definition of a prime, the only possible products that result in a prime require two of the dice to be 1 and the other a prime.  So the possibilities are :

1 1 2, 1 2 1, 2 1 1    

1 1 3, 1 3 1, 3 1 1

1 1 5, 1 5 1, 5 1 1

9 possibilities out of a total of 216 possible throws.  So probability = 9/216 

Solve (2x - 3)/6 - (4 - x)/36 = (x + 2)/9.

Hello Guest!

 \(\large \frac{2x-3}{6}-\frac{4-x}{36}=\frac{x+2}{9}\\\)

 \(\large \frac{(2x-3)6}{6\cdot 6}-\frac{4-x}{36}=\frac{(x+2)4}{9\cdot4}\\\)    | form common denominator

 \(\large \frac{12x-18}{36}-\frac{4-x}{36}=\frac{4x+8}{36}\\\)        | \(\times 36\)

 \(12x-18-(4-x)=4x+8\)  | remove the bracket

 \(12x-18-4+x=4x+8\)     | separate variables and constants

 \(12x+x-4x=8+18+4\)     | add

  \(9x=30\)                                       | divide by 9

  \(x=\frac{30}{9}=\frac{10}{3}=3\frac{1}{3}=3.\overline {33}\)

  !

I also need help plz someone help

nvm - I read the question wrong

We have:  log₈(2x),  log₄(x),  and  log₂(x)  forming a geometric progression.

Let's write these all in terms of log₂:

log₄(x)  =  log₂(x) / log₂(4)  =  log₂(x) / 2

log₈(2x)  =  log₂(2x) / log₂(8)  =  log₂(2x) / 3  =  [ log₂(2) + log₂(x) ] / 3  =  1/3 + log₂(x) / 3

So, we now have:  1/3 + log₂(x) / 3,   log₂(x) / 2,   log₂(x)

Just to make the writing simpler, let's write  log₂(x)  as  A:

--->     1/3 + A/3,   A/2,   A

To find the common ratio, we can divide the third term by the second term:

--->     (A) / (A/2)  =  2     (multiply both the numerator and denominator by 2)

We now that this ratio is also true when we divide the second term by the first term:

--->     (A/2) / (1/3 + A/3)   =   2

--->          (3A) / (2 + 2A)   =   2                  (multiplying both the numer and denom by 6)

--->                              3A  =  2(2 + 2A)     (cross multiplying)

--->                              3A  =  4 + 4A

--->                               -A  =  4

--->                                A  =  -4

log₂(x)  =  -4   --->   x  =  2^-4  =  1/16

log₄(x)  =  log₄(1/16)  =  -2

log₈(2x)  =  log₈(1/8)  =  -1

The progression is:  -1, -2, and -4

Log both sides

x log8 = (x+1) log 1/2

1.2041998 x = -.30102999

x = - .25

( but I like possible's method more!)

This might be the right way to solve this problem:

The expression "T[6]" means triangle number 6; the number to the right is the length of each side.

T[1]   --->   1

T[2] = T[1]    --->   1

T[3] = T[2]    --->   1

T[4] = T[1] + T[3]   --->   2   (add the length of a side of triangle 1 to the length of the side of triangle 3)

T[5] = T[4]   --->   2

T[6] = T[5] + T[1]   --->   3   (add the length of a side of triangle 5 to the length of the side of triangle 1)

T[7] = T[6] + T[2]   --->   5

T[8] = T[7] + T[3]   --->   6

T[9] = T[8] + T[4]   --->   8

From now on, the pattern is:  T[n]  =  T[n - 1] + T[n - 5]   ...

(tan(asin(.2))^2 = 0.0416666666666946

From omi's calcs

It looks like the floor function will result in an answer of.  -1.  0.  Or. 1

three

If you're writing a program and you have a sequence of numbers, try using a loop; usually, FOR loops are the easiest to write.

Or, do you mean that you want user input?  Then, obviously, use INPUT statements.

(1/2)^(x + 1) = 2^(-x-1)

8^x = 2^3x

3x = -x-1

x = -1/4

You can track the probabilities using different case, i.e. Anty gets to 0 after 1 steps, 2 steps, 3 steps, and so on.  This gives us e_1 = 1/2*1 + 1/4*2 + 1/8*3 + 1/16*4 + ... By arithmetico-geometric series, e_1 = 2.  Similarly, e_2 = 1/2*2 + 1/4*3 + 1/8*4 + 1/16*5 + ... = 3, so (e_1,e_2) = (2,3).

By stars and bars, the number of possible vote counts is \(\binom{52}{5} = 2598960\)

4/25 is right!  Good job.

In triangle(ABC), you have:  AB = 24,  AC = 28.  and  BC = 9.

Since you have the three sides, use the Law of Cosines to find angle(A)

  a²  =  b² + c² - 2·b·c·cos(A)

  9²  =  28² + 24² - 2·28·24·cos(A)

Solve this for angle(A).

Now, look at triangle(ADE).

AD = 35, AE = 30, and you have angle(A) from above.

Use the Law of Cosines for this triangle.

  a²  =  b² + c² - 2·b·c·cos(A)

  a²  =  35² + 30² - 2·35·30·cos(A)

The problem should say BC = 7.

Guest's answer is correct.

If you wish to do this problem algetraicly:

(n - 1)³  =  n³ - 3n² + 3n - 1

(n - 2)³  =  n³ - 6n²​ + 12n - 8

(n - 3)³  =  n³ - 9n²​ + 27n - 27

Therefore:

(n - 1)³ + (n - 2)³ + (n - 3)³  =  [ n³ - 3n² + 3n - 1 ] + [ n³ - 6n²​ + 12n - 8 ] + [ n³ - 9n²​ + 27n - 27 ]

                                            =  3n³ - 18n²​ + 42n - 36

Since n³  =  (n - 1)³ + (n - 2)³ + (n - 3)³,

          n³  =  3n³ - 18n²​ + 42n - 36

            0  =  2n³ - 18n²​ + 42n - 36

            0  =  n³ - 9n²​ + 21n - 18

Try various factors of -18 to find an integer divisor of  n³ - 9n²​ + 21n - 18.

Sooner, or later, you will get to  n = 6.

Since  n = 6  is a divisor,  n³ - 9n²​ + 21n - 18  =  (n - 6)(n² - 3n + 3)

Since  n² - 3n + 3  cannot be factored using only integers, the only answer is:  n = 6.