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Thanks CPhill for covering up for me

AIB = 134 degrees

Angle IAB + Angle IBA + 134 = 180 degrees

Angle IAB + Angle IBA = 46 degrees

AX and BY are angle bisectors

So IAB = CAB/2

and IBA = CBA/2

(CAB+CBA)/2 = 46

Angle CAB + Angle CBA = 92

That means angle C would be 180 - 92 = 88 degrees

\(1/n(n+3)=blank/n+blank/n+3. \)

1/ [n (n + 3)] = A / n  + B /(n + 3)

Multiply through by  n (n + 3)

1 =  A ( n + 3)  + B(n)

1 = (A + B)n  + 3A

Equate terms

A + B  = 0

3A   =1  →  A =1/3

So B = -1/3

Check     

(1/3) / n   - (1/3) / ( n + 3)  =

[(1/3)(n + 3)  - (1/3) n] / (n (n + 3))

[ (1/3 n  + 1  - 1/3 n ]  / [n ( n + 3)]  = 

1 / [ n ( n + 3) ]

So

first blank = 1/3

second blank  =  -1/3

They are not equal

https://web2.0calc.com/questions/help_69801

The answer is (3,-4)

I think this might be a similar question: 

https://web2.0calc.com/questions/angle-bisectors-ax-and-by-of-triangle-abc-meet-at

Are the 2 blanks equal or not?

Find what ????

Thanks so much 

thanks CPhill

Let P = (x,y)

PA^2  + PB^2 + PC^2   =

(x - 4)^2  + ( y + 4)^2 + ( x - 3)^2 + ( y -5)^2  + (x + 1)^2 + (y-2)^2

Simplifying this we  get

3 x^2 - 12 x + 3 y^2 - 6 y + 71

3 [ x^2 - 4x + y^2 - 2y ] + 71        complete the square on x, y

3 [ x^2 -4x + 4 + y^2 -2y + 1 ]  + 71 - 12 - 3

3 [ ( x - 2)^2  + (y - 1]^2 ]  +  56

Q = (2, 1)

k  =  56

CORRECTED

                              A

12                               10                  18

                  Z                      Y

                                             8

         B                                       C

                            9.6

Since  CZ is a  bisector

Let  BZ = x  and  AZ =12 -x

So

AZ / AC  = BZ / BC

(12 -x) / 18 = x / 9.6

9.6 ( 12 -x)  = 18 x

115.2  - 9.6x = 18x

(18 + 9.6)x = 115.2

x  =115.2  / ( 18 + 9.6)   =  96/23   =  BZ  ≈ 4.17

Sorry for the trouble but i cant really solve the problem with the other problem (if that makes sense) , can you help?

Here :

https://web2.0calc.com/questions/please-help-thank-you-if-you-try_2#r1

Nice, AD  !!!!

Thanks so much CPhill, can't do it without you.

This sum if really a combination of 2 geometric series.

First geometric series: \(1\over{5}\),\(2\over{25}\),\(4\over{125}\), and so on

Second geometric series: \(2\over{5}\), \(4\over{25}\), \(8\over{125}\), and so on

use: \(a_1\over1-r\) formula: The sum of the first geometric series would be: \(1\over{5}\)\(\div\)\(3\over5\)=1/3

2nd would be: 2/5 \(\div\) 3/5 = 2/3

1/3+2/3 = 1

We still didn't add the first term: 1

So 1+1 = 2

I have been working on this problems for a long time, and still did not get the answer

I don't think the answer is right

Do you think you might of did something wrong?

Probably an easier way, but.....

Height of triangle EAD =10

(1/2) the base = 5

Slope of line through AE = 10/5 = 2

So....Let the line through AE  have  the  equation  y =2(x - 10)   =  2x -20

Slope of line through CD =  [ 10 -0 ] / [0 -20 ]  =  -1/2

So....Let the line through CD  have the equation    y = (-1/2)x + 10

Find the  intersection of these lines = x coordinate of N

2x -20  = (-1/2)x + 10

(5/2)x = 30

x =  12

The y coordinate of A  is   y = (-1/2)(10) + 10  = 5  = base of triangle formed by shaded region

 (12 -10)  = 2  =  height of triangle formed by the shaded region

Shaded region    =  (1/2) (5) * 2  =   5

Pluging in random numbers for n works, but can anyone find a better way to get the possible values of n? Thank you