CPhill

  A

      4            5

     B        D          C

                3

Since AD is a  bisector.....Let BD = x  and DC  =3 - x   and  we have that

BD/ AB  = DC / AC

x / 4 = (3 - x) / 5

5x = 4 (3 - x)

5x  = 12 - 4x

9x= 12

x =  4/3  = BD

[ ABD ] = (1/2) (4/3)(4)  =  16/6 = 8/3

[ABC ]  = (1/2)(4)(3)  = 6

[ADC ]   =  [ ABC ] -  [ ABD ]  = 6 - 8/3  =   10/3 =  3 + 1/3  =   3 (to the nearest integer )

200 =  2^3 * 5^2 * 1   =  2*2*2*5*5*1

No. of distinct  sequences  =

6! / [ No. of times 2 appears ! * No. of times 5 appears ! ]

6! / (3! * 2!)  =   60

By  " brute force " .....

Laverne  says     2   7   12    17      22       27      32

Progressive sums     9      21    38        60       87      119

Another (more mathematical )  way

First term Laverne says = 2

Let the last term she says = 2 + 5n

Sum  > 100

[ 2 + (2 + 5n) ]  * n  / 2   > 100

[ 5n + 4 ] * n  >  200

5n^2 + 4n - 200 > 0      (1)            set =  0   and solve for positive n

5n^2 + 4n - 200 = 0

n = [ -4 + sqrt [ 16 + 4000] ] / 10    ≈  5.9

≈ 5.9 is a root of 5n^2 + 4n - 200....so....the next largest integer also  makes (1)  true

n = 6

And the number that Laverne says  is  2 + 5(6)  =    32

Not great at "counting" probs....here's my best attempt....corrections welcome !!!

1 3      5 6 7 8 9 10 11                7

2 4      6 7 8 9 10 11 12              7

3 5      7 8 9 10 11 12                 6

4 6      8 9 10 11 12 1                 6

5 7      9 10 11 12 1  2                6

6 8     10 11 12 1 2  3                 6

7 9     11 12 1  2  3  4                 6

8 10    12 1 2 3 4 5                     6

9 11     1 2 3 4 5 6                      6

10 12   2 3 4 5 6 7                      6

11  1    4 5 6 7 8                         5

12  2    5 6 7 8 9                         5

2(5) + 8(6) + 2(7)  =  72 ways

  See here

There are no students  only  in the Math club

3 Students are in  both  - 3 who are in the Science club are also in the Math club

There are 3 students  only in the  Science club

tan (7pi/4)

Note that this  really     tan ( 2pi - pi/4)  

So...it's just  a 45°  angle in the 4th quadrant =   tan (- pi / 4)

tan (pi/4)  =  1

But in the 4th quadrant the tangent  is  -

So

tan (7pi/4)  = tan (- pi /4)  =   -1

OPO  = (3/5)(2/4)(2/3)  = 1/5

POP  = (2/5)(3/4)(1/3)  =  1/10

1/5 + 1/10  =   3 / 10

1 step  =  4 points

2 steps  = 9 points

3 steps = 16 points

n steps  = (n + 1)^2  points

12 steps  = (12 + 1)^2  =  169 points

If you drove at 50 mph, you would have arrived  EARLIER, not, LATER

a1 + a3 + a5  =  7

a1 + (a1 + 2d) +( a1 + 4d)  =  7

3a1 + 6d  = 7

a2 + a4  = 6

(a1 + d)   + ( a1 + 3d )  =  6

2a1 + 4d  = 6

3a1 + 6d   =  7    mult through by 4

2a1 + 4d = 6      mult through by -6

12a1 + 24d  = 28

-12a1 - 24d  = -36         adding these we get

0  =  -  8

No solution

W               A   15      X

                                 m

                                 C

                                3m                            

Z         B      20         Y

tan  CAX =  m / 15   

tan  CBY  = 3m/ 20

Since angle ACB   = 60   then   angles CAX + CBY = 120

So

tan  ( CAX + CBY)  =   tan (120)

Identity  

tan (A + B)  =  [ tan A + tan B ] / [ 1 - tan A * tan B ] 

[ m/15 + (3m) / 20 ] / [ 1 - (m/15) [ (3m) / 20] ] =       -sqrt 3

Solving this  for positive m produces  m ≈  18.05  .....3m ≈  54.15

AC =  sqrt [ 15^2 + 18.05^2 ] ≈  23.47

BC =  sqrt [ 20^2 + 54.15^2 ] ≈ 57.73

Law of Cosines

AB  =  sqrt [ 23.47^2 + 57.73^2  - 2 ( 23.47 * 57.73)(1/2) ]  ≈  50.29

No

Because AD is parallel to BE.....then  because CE  is a transversal cutting  parallel lines, angle ADC  = angle BEC

But angle ACD  >  angle BCE

So  in the two triangles

ADC + ACD  + CAD  =  BEC  + BCE +  CBE

ACD  +  CAD  =  BCE  + CBE

Since  angle ACD  > angle BCE, then angle  CAD  must be  <  angle CBE

The only way that angle CAB = angle CBE  is  if A is on  BC

2T + 1B =  3.15     →  2T   = 3.15   - 1B    →   [3.15 - B ]  / 2  =  T         (1)

1M + 1B  = 3.50  →   M = 3.50 - B          (2)

1T + 2B + 3M  = 8.15     (3)

Sub (1) and (2)  into (3)

[ 3.15 - B ] / 2  +  2B  + 3 [ 3.50 - B ]  = 8.15               mutiply through by 2

[3.15 -  B ] + 4B  + 6 [ 3.50 -B ] = 16.30

-3B + 3.15 + 21  = 16.30

-3B  =  16.30 - 3.15 - 21

-3B = -7.85

B =  7.85 / 3  ≈  $2.62 = 262 cents

https://web2.0calc.com/questions/help-me_40934

6000 / 2000  =  x  /40

3 = x /40

120  = x  = 120 min

a2 + a4 + a6 + a8 + a10 + a12  = 1

a1 + a3  + a5 + a7 + a9 + a11  = -2          subtract these and we  get

Note :   a_n - a_{n -1}  =  d

d + d + d + d + d + d= 3

6d  =3

d = 1/2

Sum of terms  from  a1 to an  =  n * a1 +  (n)(n-1) / 2  *  d

S12  =  

12a1  +  66d =  3

12a1 +  66 (1/2)  =  3

12a1 + 33  = 3

12a1 =  -30

a1 = -30 / 12 =   - 5/2

Sum of an arithmetic series  from  a1 to an  inclusive =   n * a1  +  (n)(n-1) / 2 * d

S5 =      5a1 + 10d =   1/5         (1)

S10 =   10a1 + 45d =  1/10      (2)

Multiply  (1)  through by -2   and add it to (2)

25d = -3 /10

d =  -3 / 250

5a1 + 10 (-3/250)  = 1/5

5a1 + -30/250  = 1/5

5a1  = 1/5 + 3/25

5a1 = 8/25

a1 = 8/125

S15 =    15 ( a1)  +  105 d =    15(8/125) + 105 ( -3/250)   =  -3/10

26 * r^3 = 12

r^3 = 26 /12  = 13 / 6

12 * r^6 =       32nd term

12 * (r^3)^2  =  

12 ( 13/6)^2  = 

12 ( 169  / 36 )  =

169 /  3  

We can solve this

[100 + x ]  / [ 60 + x] =  [ 165 + x ] / [100 + x]

[100 + x]^2 = [ 60 + x ] [ 165 + x ]

x^2 + 200x + 10000 =  x^2 + 225x + 9900

10000 - 9900  =  25x

100 =  25x

x =  100 /  25  =  4

Check

104 / 64 = 13 / 8

169 / 104  =  13 / 8

r  =13 / 8