CalTheGreat

? I didn't even know that this was posted before...........

AC^2  =  2^2  + 3^2 - 2(3)(2)cos(ABC)  

AC^2  = 13 -12cos(ABC)  

 cos(ADC) =-cos(ABC)

We get...

AC^2 = 6^2  + 10^2 - 2(6)(10)cos(ADC)

AC^2 = 6^2 + 10^2+120cos(ABC)   

AC^2 = 136 + 120cos(ABC)   (2)

Use the subtraction method to get...

0=123+(120+12) cos(ABC)

0=123+132cos(ABC)

DB^2  = 6^2+2^2-2(6)(2)cos(DAB)

DB^2  = 40-24cos(DAB)

Therefore, DAB is obtuse

Second step!

DB^2 =10^2 + 3^2  - 2(10)(3)cos(DCB)

DB^2 =109 + 60cos(DAB)

Use the subtraction method again...

0 = 69+(60 + 24)cos(DAB)

-23/28 = cos(DAB)

We get that sin (DAB)  = √ [ 1 - (23/28)^2 ]  = √(255)/28 = sin PCB

(PB + AB) * PB = PC(PC+CD)

(PB + 2) * PB= 7/11)PB((7/11)PB + 10)

PB^2 + 2PB=(49/121)PB^2 + (70/11)PB

Now, let

PB  = x

x^2+2x = (49/121)x^2 +(70/11)x

(121 - 49) x^2 / 121 + (2 - 70/11)x  = 0

(72/121)x^2  - (48/11)x  = 0

(72/11)x = 48

So the answer is 616/9

THAT IS ONE TOUGH PROBLEM!!!!

Nice, Chris!!!

Yes, they will!

Also, posting these questions on the college math forum will help a lot!!!

Yes, it is! Nice, CU!

I'll do the 14th percentile as an example:

Using a standard normal table, the z value that corresponds to an are of 0.14 is -1.08 or z = -1.08

The corresponding x value is

x = z*σ + µ = -1.08*5 + 41 = -5.4 + 41 = 35.6

Now use this to find the 5th percentile. Hope this helped!

LOL! Thanks!!!

"Help coronavirus..."

Doesn't that mean to not go outside, use social distancing, and stop the coronavirus? I don't know what you mean by graphing...

Since a^2+b^2=c^2 or c^2-b^2-a^2

Let's say that a=9

                      c=41.

41^2-9^2= 1600

Now, square root 1600, which will get you 40.

Therefore, 40 is a side length that will form a right triangle.

Hope this helped!