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Note  that   triangles  AMC  and NBC  are  both right

And note  that CM  =(1/2) BC    and   CN  =(1/2) AC

So.....we  have  this ststem of equationa

CM^2  + AC^2  = AM^2

CN^2  + (BC)^2  = BN^2        subbing  in, we have

[ (1/2) BC]^2  + AC^2  =  19^2          simplify

[ (1/2)AC]^2  + BC^2  = 13^2

BC^2/4  + AC^2  = 19^2

AC^2/4  + BC^2 = 13^2

BC^2/4 + AC^2  = 361    ⇒  BC^2  + 4AC^2  = 1441 ⇒  -BC^2  - 4AC^2 = -1444    (1)

AC^2/4  + BC^2  = 169    (2)

Add (1) and (2)   and  we  have  that

-4AC^2 + AC^2/4  = -1275

(-15/4)AC^2  = -1595        mutiply  both sides  by  -4/15

AC^2  = 340

And  

340/4  + BC^2  =  169

340 + 4BC^2  =  676

4BC^2  = 336

BC^2 = 84

So the hypotenuse AB   can be  found as

AC^2  + BC^2  = AB^2

340  +  84  =  AB^2

424  =  AB^2

√424  =  AB

√ [ 4 * 106 ] =  AB

2√106  =  AB ≈  20.59

See  the image here :

A median bisects the side in half(1/2).

Let the side AC=2a and CB=2B, thus, the medians divide the sides into a and b, respectively.

Systems of equations gives us...

a^2+(2b)^2=(13)^2

(2a)^2+b^2=(19)^2

a^2+4b^2=169

4a^2+b^2=361

Solve for 2a, and 2b 2\sqrt 85 and b=2\sqrt21

c^2=424 ...solve for c

probably made a mistake somewheree !

.

The common denominator is:  (x + 5)(x-3)² 

Multiplying  A/(x + 5)  by  (x - 3)² / (x - 3)²  gives a numerator of  Ax² - 6Ax + 9A

Multiplying  B/(x - 3)  by  (x + 5)(x - 3) / (x + 5)(x - 3)  gives a numerator of Bx² + 2Bx - 15 B

Multiplying  C/(x - 3)²  by (x + 5) / (x + 5)  gives a numerator of  Cx + 5C

Adding these numerators together, and rearranging, we have:  (A + B)x² + (-A + 3B + C)x + (9A - 15B + 5C)

Since the term on the left-side of the equation has no  x²  term:              A + B  =  0.

Since the term on the left-side of the equation has no  x  term:      -A + 3B + C  =  0

Since the constant on the left-side equals 1:                               9A - 15B + 5C  =  1

Solving these three simulaneous equations, we get  A  =  1/64     B  =  -1/64     C  =  1/8

(I used matrices)

Group 1: 1

Group 2: 2

First one: 2

Second one: 1

Third one: 2

Angle(M) = ½·arc(ZN)   --->   Angle(M)  =  40^o 

Angle(N)  = ½·arc(XM)   --->   Angle(N)  =  17.5^o  

In Triangle(MYN), Angle(MYN)  =  180^o -  40^o -  17.5^o  = 122.5^o.

Angle(MYN) and Angle(XYZ) ae vetical angles; therefore, Angl(XYZ)  =  ....

Good to see you, EP   !!!!!

cos x

_________          multiply top/bottom by  1 + sin x

1  -  sin x

cos x  ( 1 + sin x)

___________________

(1 - sin x) ( 1 + sin x)

cos x ( 1 + sin x )

__________________

 1 - sin ^2 x

cos x ( 1 + sin x)

_____________

    cos^2  x

1 + sin x

________

  cos x

   1                sin x

_____    +   _____

cos x            cos x

sec x  +  tan x

The arc is a fraction of the total circumference:  85 / 360

TOTAL circumference =  2  pi  r

arc length  =   85/360  x   2 pi (8)     cm 

          A

 13                 15    

B       14                C

Let  B  = (0,0)   and C  = ( 14,0)

So......M  =  (7,0)  ....so   BM  =  7

Using  the Law  of Cosines

AC^2 = AB^2  + BC^2  - 2(AB * BC) cos (ABC)

15^2  =  13^2  + 14^2  - 2(13 * 14)  cos (ABC)

[ 15^2  - 13^2  - 14^2 ] / [ -2 ( 13 * 14) ]  = cos ABC

5/13  =  cos ABC

And applying this again we have  that

AM  = √ [ BM^2 + BA^2  - 2 (BM * BA) cos ABC ]

AM = √ [ 7^2  + 13^2 - 2 ( 7 * 13) (5/13) ]

AM  = √[ 49 + 169 - 2*7*5 ]

AM = √ [ 49 + 169 - 70]

AM = √148  = 2√37  units

W = F x d     joules

W = 4.0  x  55  = ......... j

A line passes through point (3, -4) and has a slope of 3.
Write an equation in Ax+By=C form for this line.
Use integers for A, B, and C.​

See  my answer  here

https://web2.0calc.com/questions/help-asap-plz_13#r1

Uh, oh ...I didn't  see the negative sign  !!!!

See my  revised answer  

thx so much!!!

Note  that  LOM  will  be  an inscribed  angle  in the smaller circle

Since  its measure  = 75°....the  minor arc  LM in the smaller circle  will measure  twice this  = 150°

So....arc LOM in the smaller circle  =  360  - 150  =   210°

What if there are more than 1 choice provided

I'm using x  instead of theta.......same difference...

3 cot x   = -√3      divide  both sides  by  3

cot  x = -√3 / 3  =    -1   / √3

Note  that  this is  the  same  as 

tan x  = - √3

And this  happens  at

x =   (2pi)/3  +  pi * n

EDITED ANSWER    !!! 

Its  is correct

It's   means  " it is"

I like to do it  this way with just  one  variable.....this eliminates  setting up two equations.....

Let x  be  the  number  of  apples  sold  and  20 - x  be the number of bananas sold....so....we have this equation

.50x  + .75 ( 20 - x)  = 11.50    simplify

.50x  + 15 - .75 x =  11.50         subtract 15  from both sides  and simplify

-.25x = -3.50       divide  both sides  by  .25

x = 14  = apples sold

And

20 - 14  =  6  = bananas sold

One  of  the white  sectors  must  comprise  1/21  of the area of the  circle

So

13pi/54 =  pi * r^2   / 21       multiply both sides by  21    and divide out pi

(21 * 13)  / 54  = r^2

91/18 = r^2       take  the positive root

√ [ 91 / 18 ]   ≈   2.25  in  =   r

x^2 + y^2-2x-2y-2=0 and x^2+y^2+2x+2y-2=0

Setting  these  equal  we  get that

x^2 + y^2-2x-2y-2  =   x^2+y^2+2x+2y-2       simplify

-2x - 2y   =  2x + 2y

-4x  = 4 y

-x  =  y

Sub  this  into  eithe equation for  y

x^2  + (-x)^2  - 2x - 2(-x) - 2  =  0

x^2 + x^2  - 2x + 2x  -  2  =  0

2x^2  - 2  =  0

2x^2  =  2

x^2  =  1             take  both roots  and  we have that

x = 1       and  x  =  - 1

And when x  =1,  y  =-1

And when x  = -1, then y = 1

So  the points of intersection   are  ( -1, 1)  and (1 , -1)

See  the  graphs here :  https://www.desmos.com/calculator/rjkmzcqniv

Thank You!