All Answers 356371 Answers

Distance = Rate x Time

Runner B ran 75 miles.

You can start from 999 and continue down until you found a number that meets the requirment. 999 wouldn't work because when divided by 3, there is no remander. I'm sure it will not take too long.

k takes on the following value: 1,  2,  4,  5  for which 3x mod 6 ==k, has no solutions.

Using "CRT + MMI", we have:

(4 x) mod 20 = 8
(3 x) mod 20 = 16, solve for x

x = 20m + 12, where m = 0, 1, 2, 3.........etc.

x =20 * 1  +  12 = 32

x^2 =32^2 mod 20 == 4 - The remainder.

I dont understand your edited question.....

first you put 3(18)=?

Then you edited it to Log 3(18)= 3 How?

A = [3*(3 / tan60º + 3 / tan30º)] / 2 = 6√3

3(18)=54

Let the number of apples  = 6a

Let the multiple of n  =  bn

So

6a  - 1  =  bn 

6a  = bn + 1

6a  will  always  be an even integer

Then  n  cannot  be even   because   then   bn + 1  would be odd 

So.....n must  be odd =  1,3,5,7,9

How many different 4-digit numbers have exactly two different digits?

There are 576 four digit numbers that have exactly 2 different digits!

ab   =10

a^2  + b^2  = 40

( a + b)^4 =  a^4  + 4a^3b  + 6a^2b^2  + 4ab^3  + b^4     (1)

(a  - b)^4  =  a^4  - 4a^3b  +  6a^2b^2  -  4ab^3  + b^4    (2)

Subtract  (2)  from (1)  and we get that

8a^3b + 8ab^3  =    8(ab)( a^2  + b^2)

Now....just fill in what we  know

8 (ab) (a ^2 + b^2)   =

8 (10) (40)  =  

3200

tysm for you help!

The sum of the interior angles   =  540°

So

(x + 1)  + 2x  + 3x + 4x + (5x -1)  =540      simplify

x + 2x + 3x + 4x + 5x   =540

15x  =  540   divide both sides by 15 

x =36°

Largest angle  =  5 (36)  -1  =   179°

We have a 30-60-90  right triangle

You  should memorize  these relationships in a  a 30-60-90  triangle

30          60              90

x      x* sqrt (3)        2x

The  angle opposite  the 30° angle  =  x =   6

The  angle opp  the  60°  = x* sqrt (3)  =   6*sqrt (3)  = DE

The angle opp the 90° angle  = 2x  =   2 (6)    =12  = DF

So

DE + DF  = 

6sqrt (3)  + 12

thx so much

x / (x +4) =  -9 /( x + 5)     cross-multiply

x ( x + 5)  = -9 ( x + 4)     simplify

x^2 + 5x   =  -9x  - 36         rearrange as

x^2  + 14x + 36  =  0       complete the square on x

x^2  + 14x   =   -36           take 1/2  of 14  =7.....square this =  49....add to both sides

x^2 + 14x + 49  = -36 + 49            factor the left, simplify the right

(x + 7)^2  =  13            take both roots

x+ 7  = sqrt (13)     or        x + 7   =  -sqrt (13)

x = sqrt (13) - 7        or       x  =  -sqrt (13)  - 7

You're welcome   ....

See the following :

Using the Law of Cosines   we  can find EF = one side of a square  which comprises part of the shaded region

EF  =  sqrt  [ BE^2  + BF^2  - 2 ( BE * BF)cos (30°)  ]

EF  = sqrt [  1^2 + 1^2  - 2 (1 * 1)sqrt (3/2) ] 

EF = sqrt  ( 2 - sqrt (3) )

EF^2   = area of the square =  (2 -sqrt(3) )

And the area  between the circle and the side of the square  =  

Area of  30°  sector  of  the circle =  (1/12) pi 1^2  =  pi/12      less

Area of triangle BEF  =(1/2) (1)^2 sin (30°)  =  (1/2) * (1/2)  =  1/4

So this area  =  pi/12 - 1/4   =   [ pi - 3] / 12

And we have 4 of these regions =   [pi -3 ] / 3 =  pi/3 -1

So....the total  shaded area  =   (2 -sqrt (3))  + (pi/3 -1)   =

[1 + pi/3  - sqrt (3)] units^2  ≈  .315 units^2

thanks for the help!

The least positive integer n = 105

28105 mod 365 == 0

I agree! This is also a math site and this is not even a math question anyway.......

See answer #2 here:  https://web2.0calc.com/questions/please-help-question-not-answered#r2

The chord of a circle divides the circle into two parts.  A square of area 16 is inscribed in one part, and a square of area 144 is inscribed in the other part.  The radius of the circle is?

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

Note that only 2 vertices of each square are on the circle!

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

∠CBR = ∠PQO = tan^-1(CR / BR) ==> 14.03624347º

P and Q are midpoints on TS and BC

CQ = 1/2 * sqrt(CR² + BR²) = √68

PQ = 1/4(AB + CD) = 4

OQ = PQ / cos∠PQO = √17

CO = r = sqrt(OQ² + CQ²)   ==>     r = √85

Using the Law of Sines

XY / sin 60  = 4/sin45

XY   =  4sin 60  /  sin 45

XY  = 4 sqrt (3) / 2  / ( sqrt (2) / 2 )

XY  = 2*2sqrt (3) / sqrt (2)

XY =  4sqrt (3) /sqrt (2)

XY    =  4 sqrt (6)  /2

XY  = 2sqrt (6)

The  common ratio  is   18 /12  =   3/2

And we can solve this

(first term) (common ratio)^ ( 3  -1)   = 12

(first term ) (3/2)^(3 - 1)   =  12

(first term) ( 3/2)^2  = 12

(first term) (9/4)   = 12

first term  =  12  (4/9)  =  48/9   =  16/3