All Answers 355926 Answers

real solutions would be x=5, x=15. complex solutions would be x=10-5i, x=10+5i

Hint: Use the binomial theorem. 

Vertex occurs at -b/2a which is at x=9/2 and y=51/2.

Take the conjugate to get 2021-2022i.

Simple calculations get 640 (Just list out all of the factors and add! )

Because 2 and 7 are relatively prime, we can divide the congruence to get 2n=3 mod 7, and solving gets us n=5 mod 7. So, it is simply 5,12,19,26,33,40,47,54 for a total of 8 numbers.

Units digit of \(3^{4k}\) is always 1 so the units digit of this is 2. 

Also I think 2^2^1000 +1 is what you mean, if that is it, then the units digit of \(2^{4k}\) is always 6 so the units digit of this is 7

Factor is to get \((x-5)(x-15)(x^2-20x+125) \) which has solutions of \(x=5,x=15,x=10-5i,x=10+5i\)

Complementary counting gets 1-(1/2)^4 (probabilty all are odd) = 15/16

Sum = 2: 1,1

Sum = 3: 1,2 or 2,1

Total: 36

So 3/36 or 1/12

(1/2)^6 = 1/64

\(4x^6+4x^3+1+7x^4-5x^3-6x-2 = 4x^6+7x^4-x^3-6x-1\)

Probabilty of none: \(\frac67 *\frac78 * \frac 89 * \frac{9}{10} = 0.6\)

So probabilty of at least 1 is 0.4 or 2/5

Equation of line: y=0.5x+2

x intercept: x=-4

y intercept: y=2

so the area is 4

The number of arrangements is 9!/(2!*3!*4!) = 1260.

 24-(2*3)=

18  ways

The maximum value of x - y is 3*sqrt(2).

The answer is 14.

The shortest possible distance of the trip is pi.

The probability is 8/13.

What?

The side lengths of a cyclic quadrilateral ABCD are provided in the diagram. Find BD^2.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

a = 1     b = 2     c = 3     d = 4      s = 5

Using the above formula, we can calculate the circumradius R.

R = 2.002602473

Now, we can calculate the measure of angle C.

Angle C = 101.53695902º

Finally, using the law of cosines, we can find the length of BD.

BD ≈ 3.924         BD² = 15.4

Let's  suppose   that  we  can let anyone  stand  anywhere.....there  are 4!  = 24  ways to do this

If Ben  and Amir DO  stand next to each other  they can occupy  any 3 postions  and be  arranged in 2  ways  for  each of  these   =    3 * 2  =  6 ways

So....the  number of ways  they can  be arranged  where  they ARE NOT standing next to each other  =

24  -  6    =    18  ways

We can arrange the  Astronomy books in 4 !  =  24 ways

We can arrange  the Biology books in 3 !  = 6  ways

We can arrrange  the Chemistry books in  2  ways

And we  can arrange the  subjects in 3 !  ways = 6

So

24 * 6  * 2  * 6   =   1728 different ways