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Well if the 2 angles are 30 and 60 degrees, this has to be a 30-60-90 right triangle.

Heres a pic i made:

Note that angle CEA = angle DCA + angle DAC and angle BDE = angle DCE + angle EDC, so angle DCA = angle CEA - angle DAC and angle DCE = angle EDC - angle CED.

Also, angle BDE = angle ABD + angle ADB = angle AEC + angle DAC, so angle DCE + angle DCA = (angle EDC - angle CED) + angle DCA.

Therefore, angle ACE = angle ACD + angle DCE = angle CEA, which implies that triangle ACE is isosceles.

Note that angle CEA = angle DCA + angle DAC and angle BDE = angle DCE + angle EDC, so angle DCA = angle CEA - angle DAC and angle DCE = angle EDC - angle CED.

Also, angle BDE = angle ABD + angle ADB = angle AEC + angle DAC, so angle DCE + angle DCA = (angle EDC - angle CED) + angle DCA.

Therefore, angle ACE = angle ACD + angle DCE = angle CEA, which implies that triangle ACE is isosceles.

Note that angle CEA = angle DCA + angle DAC and angle BDE = angle DCE + angle EDC, so angle DCA = angle CEA - angle DAC and angle DCE = angle EDC - angle CED.

Also, angle BDE = angle ABD + angle ADB = angle AEC + angle DAC, so angle DCE + angle DCA = (angle EDC - angle CED) + angle DCA.

Therefore, angle ACE = angle ACD + angle DCE = angle CEA, which implies that triangle ACE is isosceles.

This is quite complicated but if u solve this system of equations i think you get: \(x=\frac{14}{3},\:z=\frac{8}{3},\:y=\frac{14}{3},\:L=\frac{26}{3}\)

\(\frac{14}{3}+\frac{8}{3}=\frac{22}{3}\)

The slope intercept form is: \(y=mx+b\)

First you need to find the slope (m). \(m=\frac{y_2−y_1}{x_2-x_1}.\)

\(m=\frac{−2−\left(−5\right)}{6−\left(−3\right)}\)

\(m=\frac13\)

Next, you find the y-intercept, or b. You can plug in any of the 2 points like this: \(y=\frac13x+b\)

\(-2=\frac13\cdot6+b\)

\(2+b=-2\)

\(b=-4\)

So our final answer is \(y=\frac13x-4\)

(I just wanted to make ElectricPavlov's solution more clear)

HINT: There are 4 balls and 3 boxes, where all the balls are different. One of the 3 boxes has to have 2 balls.

The point will be of the form    x, 0

Use the distance formula between this point and EACH of the given points and equate them

sqrt [(x-2)^2 + 0^2]     = sqrt [  (x-0)^2  + (0-6)^2 ]             square both sides and simplify to

        x^2 -4x +4          =            x^2 + 36                            now subtract x^2 from both sides and continue to solve for 'x' in the coordinate pair (x,0)

I believe ur question is uncomplete.

Simple interest is generally the interest rate per year without compounding.

Take the interest rate in decimal form      multiply by the number of years     and  multiply by the deposit amount to get the interest earned.

Example :    1500 deposit    10% interest   4 years        .10  *  4  *  1500

3x^2-24x+72               re-arrange

3 (x^2 -8x) + 72          complete the square for 'x'   

3 ( x-4)^2  -48 + 72

3 (x-4)^2 +24                now add a b c

See if this helps:     Interior angle for a polygon =  ( n-2) *180 /n    for octagon this equals 135^o

   EXTERIOR angle for the octogon is then  180 - 135 = 45^o

4x^2 + 15x + 2 .

4 (x^2 + 15/4 x) + 2         'complete the square for x'

4 (x+ 15/8)^2  + 2  - 4 * 225/64

4 ( x+ 15/8)²  - 12  1/16                          

Sub in x = 1

a - 6 + b -5 = 5 

a+b - 16 = 0    (1)

sub in x = 2

8a - 24 + 2b -5 = -53

8a +2b +24 = 0        (2)

Two equations with two unknowns   solve for a and b

mutilpy (1) by -8

-8a-8b+128 = 0    add to (2)

-6b +152 = 0

b = 25 1/3      a= - 9 1/3

$ 2x^2+6x+11  $

$ 2\left(x^2+3x+\frac{11}{2}\right)  $

$ 2n=3 \implies n=\frac{3}{2} $ which you want to square $ \implies \left(\frac{3}{2}\right)^2 $

$ 2\left(x^2+3x+\frac{11}{2}+\left(\frac{3}{2}\right)^2\right) =2 \left( \left(\frac{3}{2}\right)^2 \right) $

$ 2\left(x^2+3x+\frac{11}{2}+\left(\frac{3}{2}\right)^2-\left(\frac{3}{2}\right)^2\right)  $

completing the square 

$ 2\left(\left(x+\frac{3}{2}\right)^2+\frac{11}{2}-\left(\frac{3}{2}\right)^2\right)   \implies  2\left(\left(x+\frac{3}{2}\right)^2\right) + 2\left( \frac{11}{2}-\frac{9}{4} \right) $

           $\Updownarrow   $

$ 2\left(x+\frac{3}{2}\right)^2+\frac{26}{4} \equiv a(x - h)^2 + k $ 

$ \overset{. \: .}{\smile } $

2 (x^2 +3x)   = -11

2 (x+1.5)^2 = -11 + 2.25       You can finish, right?

Multiply all terms by \(x^{2/3}\)  (and remember that the cube root of x is the same as x^1/3).

Make use of the fact that \(x^a\times x^b=x^{a+b} \)

Collect like terms together.

Solve the resulting equation for x.

For (a) simply substitute the values x = 5 and y = 12 into both expressions and calculate the results.

For (b)  set (x+y)² = x² + y² then expand the left hand side and simplify the resulting equation.  You will then have a condition that must be satisfied for the two ex[ressions to be equal.

The following should help you decide:

good spotting Catmg    

 Thanks for the info i will try to figure it out for more   
 

7!/(3!*2!*2!)

Arc RG = 360 - (arc FR + arc FG)

Angle RPG = 1/2(arc RG + arc FQ)