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A = Arman

B = Babken

C = Cecilia

B = A - 10

C = B - 10

C = A - 10 - 10 = A - 20
C = 100 - 20 = 80m

When Arman was at the finish line 100m away, Cecilia ran 80m, which is 20m away from the finish line.

Sum{18, 19, 20, 21, 22} = 100

mharrigen920,

Why did you vote geno down?

I suppose this is your standard response when people invest their time and expertise in helping you.

Solve by squaring both sides:

                                                      sqrt(3)·cos(x)  =  sin(x) + 1

                                                   sqrt(3)·cos(x) ]²  =  [ sin(x) + 1 ]²  

                                                              3·cos²(x)  =  sin²(x) + 2·sin(x) + 1

Since  cos²(x)  =  1 - sin²(x)

                                                       3·( 1 - sin²(x) )  =  sin²(x) + 2·sin(x) + 1

                                                            3 - 3·sin²(x)  =  sin²(x) + 2·sin(x) + 1

                                                                             0  =  4sin²(x) + 2sin(x) - 2

                                                                             0  =  2sin²(x) + sin(x) - 1

                                                                             0  =  ( 2sin(x) - 1 )( sin(x) + 1 )

So:  either       2sin(x) - 1  =  0       or     sin(x) + 1  =  0

                             2sin(x)  =  1                     sin(x)  =  -1

                                sin(x)  =  ½                          x  =  sin^-1(1)

                                       x  =  sin^-1( ½  )

Now, you'll need to finish this ...

I get everything up until the last part, in the parenthesis. What do you mean 14pi/3 has to be reduced into an angle smaller than 2pi? How would I do that? 

Hint for #1:

cubic polynomial:  f(x)  =  a·x³ + b·x² + c·x + d

                             f(1)  =  a·(1)³ + b·(1)² + c·(1) + d  =  9

                             f(1)  =  a + b + c + d  =  9

Since each of the coefficients must be a positive whole number, the values for a, b, c, and d must be:

(in some order)  1, 1, 1, 6

                           1, 1, 2, 5

                           1, 1, 3, 4

                           1, 2, 2, 4

                           1, 2, 3, 3

                           2, 2, 2, 3

Now, the problem is to find how many different ways each of the above can written.

First:  you will need to place the complex number   -2·sqrt(3) - 2i     into  r·cis(theta)  form.

For the complex number   a + bi     r  =  sqrt( a² + b² )     and     theta  =  tan^-1( b/a)

For your problem:  a  =  -2·sqrt(3)   and   b  =  -2:

     r  =  sqrt( ( -2·sqrt(3) )² + (-2)² )  =  sqrt( 12 + 4 )  =  sqrt( 16 )  =  4

     theta  =  tan^-1( -2 / (-2·sqrt(3) )  =  tan^-1( sqrt(3)/3 )  =  7pi/6

r·cis(theta)  =  4·cis( 7pi/6 )

General rule:  [ r·cis(theta) ]ⁿ  =  rⁿ·cis( n·theta )

Now, to find the 4^th power, take the r-value to the 4^th power, and multiply the angle by 4.

[ 4·cis( 7pi/6 ) ]⁴  =  4⁴·cis( 4·7pi/6 )  =  256·cis( 14pi/3​ ) 

(You'll have to reduce the 14pi/3  into an angle smaller than 2pi.)

For the complex number  a + bi,  

the modulus is  sqrt( a² + b² )

and the argument is the tan^-1( b/a )    [placed in the correct quadrant]..

2 + 2i:     modulus  =  sqrt( 2² + 2² )  =  sqrt( 8 )         

               argument  =  tan^-1( 2 / 2 )  =   tan^-1( 1 )  =  45^o.

4i  =  0 + 4i:   modulus  =  sqrt( 0² + 4² )  =  4

                      argument  =  tan^-1( 4 / 0 )  =  90^o.

Owen's tosses could be:

WWW     *

WWL      *

WLW      *

WLL       *

LWW      *

LWL       *

LLW       *

LLL

Of the 8 possibilities, 7 have at least one winner.

We need to consider only these 7 possibilities.

One of these seven has all three winners     --->     Probability  =  1/7

We have right triangle(ABC) with angle(B) the right angle.

AB = 6     N is the midpoint of AB, so BN = 3.

Triangle(NBC) is also a right triangle, with sides BN = 3 and BC = 3sqt(3).

By the Pythagorean Theorem, CN²  =  BN² + BC² 

                                                 CN²  =  (3)² + ( 3sqrt(3) )²

                                                 CN²  =  9 + 27  =  36

                                                 CN  =  6

The point P is the centroid of the triangle, so CP = 2/3^rds of CN.

This makes CP = 4.

385  =  5 x 7 x 11

So, two sides will be              5 x 7    =  35    x 2  =  ...

another two sides will be       5 x 11  =  55    x 2  =  ...

and the last two sides will be 7 x 11  =  77    x 2  =  ...

You'll need to finish the multiplying and adding the answers together ...

As far as I can see, there is NO integer solution. 2, 8 or 3, 7 or 4, 6. None of these will balance the 2nd equation, unless it is 52, or 58 instead of 56.

Check https://web2.0calc.com/questions/help-asap-please_34.

Try this: [6 - 1] C [4 - 1] = 5 C 3 =10 ways for gummy bears.

              [6 - 1] C [4 - 1] =5 C 3  =10 ways for chocolates

So, total number of ways = 10 x 10 = 100 ways ?

A=abs((30*6) - (5.5*48)), solve for A

A = 84 Degrees.

Since the two numbers are consecutive integers, they will be very close to the square root of 1980.

sqrt(1980)  =  44.508... 

So, try 44 and 45.

Solve the quadratic x^2 - 10x + 9 = 0. 

Alternatively, this one is simple enought that you could just factor it. 

What two numbers multiply together to get +9 and add together to get –10 ? 

I wish they were all this easy.  Of course, the numbers are –1 and –9 

So,   x² – 10x + 9 = 0   factored is   (x –1)(x – 9) = 0  

Setting each component equal to zero:    (x – 1) = 0 

                                                                 x = 1 

                                                                 (x – 9) = 0  

                                                                 x = 9 

_.

thats wrong.

Hi Alan, thanks for responding. I tried your method as follows:

Plugging x=50 into the first equation:

y= 73+45+2

y=120

Plugging x=50 and y=120 into the second equation and solving for k:

120=2(50) + k

k=20

I tried inputting this answer but it is saying my answer is wrong. I also don't understand where you got x=50 from.

By stars and bars, the number of ways is C(6,4)*C(6,4) = 225.

thanks!! I thought it was only integers ;(

It is 10!

x +y =10,

x^2 +y^2 =56, solve for x, y

Use substitution to get:
x = 5 - sqrt(3)  AND  y = 5 + sqrt(3), OR:
x = 5 + sqrt(3)  AND  y = 5 - sqrt(3)