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m = 2j

j+m = 57       substitute first equation into second one

j + 2j = 57

3j = 57

j = 19

               then, since    m = 2j    = 38                  check:   38 + 19 = 57   correct !

x = kg/m²  /  kg/m    =   kg/m²   * m/kg  =  1/m           (see how you flip the denominator then multiply and some things 'cancel' out ?)

x = 1/1 =   1 m^-1

It will be   - b/c   = - 3/4

You have to do what's in the parentheses first.

(14+38)= 52

1316-52= 1264

Your answer is 1264

Smart

PEDMAS

1316 - ( 52) = 1264

pressure = force / area

2000 N / .68 m² = 2941.18 N/m²

= (4x40) / 5 = 160/5 = 32

listfor(n, 1, 10000, (n^2+n-1) / (n+2)! ==(1 / 6, 5 / 24, 11 / 120, 19 / 720, 29 / 5040, 41 / 40320, 11 / 72576, 71 / 3628800, 89 / 39916800, 109 / 479001600........etc==It converges to 1 / 2

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729x + 731y = 1508   (1)

725x + 727y = 1500   (2)

Subtract (2) from (1)

4x + 4y = 8  or  x + y = 2  so y = 2 - x;

725*x + 727*(2 - x) = 1500

727*2 - 2x = 1500

2x  = 727*2 - 1500

x = -23

y = 2 - (-23) = 25

x - y = -23 -25 = -48

Edited to correct silly mistake!

Thank you so much, CPhill!

Thanks very much OldTimer.

And a merry Christmas to you too!

(28, 98, 168, 238, 308, 378, 448, 518, 588, 658, 728, 798, 868, 938, 1008, 1078, 1148, 1218, 1288, 1358, 1428, 1498, 1568, 1638, 1708, 1778, 1848, 1918, 1988) >>Total = 29 such numbers.

"Mean score of 6 games is 15. The mean game of 7 games is 16. How many points were score in the seventh game?"

6 games:    15 = N/6     N = 6*15

7 games;   16 = (N+s)/7    where s is score in seventh game, so   N + s = 7*16

Hence 6*15 + s = 7*16

s = 7*16 - 6*15 = 22

15/77

This one is a little hard to  think  about....

The function x^2 - 7x + 13   will  never   =  0   since it lies wholly above the  x axis

Since we  cant  divide  by  0, we actually  want to find the  x values  where 

x^2 -7x + 13  =  1

The two x values that  make this true will produce  a floor value of  1

The  two solutions that we  find will be  included  in the  domain....all real numbers  between these WILL NOT be in the  domain   since they  will make the floor value    = 0    and we  can't  divide  by  that

So  rearranging, we have 

x^2 -7x + 12   = 0     factor

(x - 3) ( x - 4)   =  0

Setting each factor to   0  and  solving for x produces  x  =3    and  x  = 4

So....the domain will be  (-inf. 3 ] U  [4, inf )

This graph seems to  confirm  our answer  : https://www.desmos.com/calculator/df9v0bh5wt

Since ADCB is a trapezoid, then AD is also parallel to BC

And note that angle  EBC   = angle BDA

And because AB is parallel to CE, then angle BEC  = angle DBA

So  triangle  BEC  is similar to triangle  DBA

So    BE  and  DB   are corresponding sides

And the ratio of the area of DBA to DEC  = (BD/ BE)^2 =  (8/15)^2  = 64/225

And   triangles EDC  and BEC  share  equal altitudes  so  their areas are to each other  as  their bases

ED = 7      BE  = 15

So    area   EDC =  (7/15)area  of triangle BEC  ⇒  EDC/ BEC =  7/15 =  105/ 225

And the ratio of the area of DBA to BEC  =  (8/15)^2  =  64/225   ⇒  DBA / BEC = 64/225

So

(DBA / BEC)         64 / 225             [ DBA ]           64

__________  =    ________  =      ______  =    ____

(CDE /  BEC)        105 / 225           [ CDE  ]        105

c / ( c - 5)  =  4 / (c - 1)     cross-multiply

c ( c - 1)  =  4(c - 5)

c^2  - c   = 4c - 20

c^2 - c - 4c + 20   =   0

c^2  - 5c  +  20    =  0         complete the square on c

c^2 - 5c + 25/4   =  - 20 + 25/4

(c - 5/2)^2   =  -55/4       take  both roots

c - 5/2   =   ± i sqrt ( 55) / 2

c =     5/2  ± i sqrt (55)  /2

I think  this is

(18 - 6x)                   6 ( 3 - x)                     -6 

___________   =   ____________   =    _________

x^2 - 9x+ 18            (x - 3) ( x - 6)            x - 6

We will have  a vertical asymptote  when    x -  6  = 0   ⇒     x   =    6

Additionally.....we  will  have a "hole"   when  x = 3  

So the domain is   all reals except  x =  3  and x  = 6

The circumcenter  is  the point where all three perpendicualr  bisectors meet  (we actually only need two of them)

In this case.......this is a right triangle  with the hypotenuse  = AC....there is an easy way to find the  circumcenter n this situation....

See the graph here :  

The circumcenter is located at the  midpoint of  hypotenuse AC  =  [ (-3 - 1)/2 , ( 3 +1)/2  ]  = (-2, 2) = D on the  graph

3) Easiest  way to do this is to evaluate  the functions as  if they WERE continuous

So

sqrt (2x  -6)    when x  =3      =  0

-(x - 1)^2  + 1  when  x  = 3    =   -3      

The  first two functions are  not  continuous when  x  =  3

And

-(x - 1)^2 + 1  when x  = -1 =  -3

-2x -3  when x = -1 =  - 1

So these two functions are  not  continuous at  x  = -1

See the graph here  :  https://www.desmos.com/calculator/n8zrrd4nbn

First die = any number
2nd die  =5/6 probability that it is different from the first.
3rd die  =4/6 probability that it is different from the first two.
4th die  = 3/6 probability that it is different from the first three.

Probability is =1 - [6/6 * 5/6 * 4/6 * 3/6]=1 - 360/1296 =936 / 1296 =13 / 18 - probability that at least 2 dice will show the same number.

\(f''(x)\ =\ \sin(\frac{x^2}{8})-2\cos(x) \)

Let's plug in each of the given critical x-values into the second derivative to see whether it is positive or negative.

\(f''(0)\ =\ \sin(\frac{0^2}{8})-2\cos(0)\ =\ \sin(0)-2\cos(0)\ =\ 0-2(1)\ =\ -2\ <\ 0\)

Since  f''(0) < 0  the graph of  f  is concave down at  x = 0  and so a local max occurs at  x = 0

\(f''(6.949)\ =\ \sin(\frac{6.949^2}{8})-2\cos(6.949)\ \approx\ -1.817 \ <\ 0\)    (assuming  x  is in radians)

Since  f''(6.949) < 0  the graph of  f  is concave down at  x = 6.949  and so a local max occurs at  x = 6.949

Here's some more info about this:

First one

a)  Limit  as  x approaches infinity   = 4

b) Limit as  x approaches 3 from the left    =  7

c) Limit as x approaches  3 from the right   =  infinity  (does not exist)

d ) Limit as x approaches -1 from both sides  ≈  -3.5

e) h(-1)    = 2