All Answers 356458 Answers

Here are three possible methods you could use, together with a graph of the triangles to help:

By the median theorem ON = QN * 1/3 = 4

NR = PR / 2 = 7

OR = sqrt( 4² + 7² ) = √65

a=5

b = ?

[(5*b - b^2/2)]  /  [2*(sqrt(25 - (b^2/4)))] =4/3, solve for b

Solve for b:
(5 b - b^2/2)/(2 sqrt(25 - b^2/4)) = 4/3

Cross multiply:
3 (5 b - b^2/2) = 8 sqrt(25 - b^2/4)

3 (5 b - b^2/2) = 8 sqrt(25 - b^2/4) is equivalent to 8 sqrt(25 - b^2/4) = 3 (5 b - b^2/2):
8 sqrt(25 - b^2/4) = 3 (5 b - b^2/2)

Raise both sides to the power of two:
64 (25 - b^2/4) = 9 (5 b - b^2/2)^2

Expand out terms of the left hand side:
1600 - 16 b^2 = 9 (5 b - b^2/2)^2

Expand out terms of the right hand side:
1600 - 16 b^2 = (9 b^4)/4 - 45 b^3 + 225 b^2

Subtract (9 b^4)/4 - 45 b^3 + 225 b^2 from both sides:
-(9 b^4)/4 + 45 b^3 - 241 b^2 + 1600 = 0

The left hand side factors into a product with five terms:
-1/4 (b - 10) (b - 8) (9 b^2 - 18 b - 80) = 0

Multiply both sides by -4:
(b - 10) (b - 8) (9 b^2 - 18 b - 80) = 0

Split into three equations:
b - 10 = 0 or b - 8 = 0 or 9 b^2 - 18 b - 80 = 0

Add 10 to both sides:
b = 10 or b - 8 = 0 or 9 b^2 - 18 b - 80 = 0

Add 8 to both sides:
b = 10 or b = 8 or 9 b^2 - 18 b - 80 = 0

Divide both sides by 9:
b = 10 or b = 8 or b^2 - 2 b - 80/9 = 0

Add 80/9 to both sides:
b = 10 or b = 8 or b^2 - 2 b = 80/9

Add 1 to both sides:
b = 10 or b = 8 or b^2 - 2 b + 1 = 89/9

Write the left hand side as a square:
b = 10 or b = 8 or (b - 1)^2 = 89/9

Take the square root of both sides:
b = 10 or b = 8 or b - 1 = sqrt(89)/3 or b - 1 = -sqrt(89)/3

Add 1 to both sides:
b = 10 or b = 8 or b = 1 + sqrt(89)/3 or b - 1 = -sqrt(89)/3

Add 1 to both sides:
b = 10 or b = 8 or b = 1 + sqrt(89)/3 or b = 1 - sqrt(89)/3

(5 b - b^2/2)/(2 sqrt(25 - b^2/4)) ⇒ (5 8 - 8^2/2)/(2 sqrt(25 - 8^2/4)) = 4/3
4/3 ⇒ 4/3 ≈ 1.33333:
So this solution is correct

(5 b - b^2/2)/(2 sqrt(25 - b^2/4)) ⇒ (5 10 - 10^2/2)/(2 sqrt(25 - 10^2/4)) = (undefined)
4/3 ⇒ 4/3 ≈ 1.33333:
So this solution is incorrect

(5 b - b^2/2)/(2 sqrt(25 - b^2/4)) ≈ -1.33333
4/3 ⇒ 4/3 ≈ 1.33333:
So this solution is incorrect

(5 b - b^2/2)/(2 sqrt(25 - b^2/4)) ≈ 1.33333
4/3 ⇒ 4/3 ≈ 1.33333:
So this solution is correct

The solution is:

 b = 8 - The 3rd side of isosceles triangle.

I also am using stars and bars method.

53C3 =   23426 ways

We have 50 people divide by 3 bars.  That will create 4 groups of people

Now we have to decide where the 3 bars go.

Hence 53C3

I have no idea what this means either.   

Power of respect  ?? Sounds like nonsense.

angle DAE = 180-126 = 54º

angle EBC = 90-48 = 42º

angle BAE 180 - (48+54) = 78º

You could use the on-line graphing capability of Desmos:  https://www.desmos.com/calculator

You would enter: y= sin(x)-cos(x) {0≤x≤2*pi}

You sound surprised.

x + y + z = 10,

2x + 2y + 2z = 20

   2x + 2y + 2z = 20

+ -2x - 2y - z = -32

__________________

z = -12

x + y - 12 = 10,

x + y = 22,

3x + 3y = 66

3x - 2y - 24 = -8,

3x - 2y = 16

   3x + 3y = 66

-  3x - 2y = 16

_________________

5y = 50

y = 10

x + 10 - 12 = 10,

x = 12

(x. y, z) = (12, 10, -12)

3^3+(2-8)^2-[(6-7)(14)+2

Hello Guest!

I close the square bracket.

\(3^3+(2-8)^2-[(6-7)(14)+2\ \color{blue}]=75 \)

  !

UVEQUALSTWELWE

1/  Chose the length of QR

2/  Find angles UPQ  and  UQP

3/  Calculate side QU of a triangle PQU

4/  Find angle VQU

5/  UV = sin(VQU) * QU = 12

z = -3 - 7i.

a^3 + b^3 + c^3 = 18.

x = 4 and  y = 8.

By stars and bars, the number of ways is C(51,4) = 249900.

K = (-11,1).

z = -5 - 2i.

Look under your first post.

3^3 + (2 - 8)^2 - (6 - 7)×14 + 2

27 + (-6)^2 - (-1 * 14) + 2

27 + 36 +14 + 2 = 79

There are 14 different positive values of x^y.

productfor(n, 1, 50, (2*n - 1) = 272539213 9750729502 9807132454 0091863329 0796330545 8034137343 2882344310 6201171875

Factor the above answer: 3^26×5^12×7^8×11^5×13^4×17^3×19^3×23^2×29^2×31^2×37×41×43×47×53×59×61×67×71×73×79×83×89×97 (81 prime factors, 24 distinct)

The largest n = 26