check the pictured below.
now, notice that APB is just a flat-line and therefore those two angles there add up to 180°.
also notice that since ∡ACB is 90°, then those two angles add up to that much.
![\bf \stackrel{\measuredangle ACP}{(3x+2y)}+\stackrel{\measuredangle BCP}{(3x+4y)}=90\implies 6x+6y=90\implies \stackrel{\textit{common factor}}{6(x+y)}=90 \\\\\\ x+y=\cfrac{90}{6}\implies x+y=15\implies \boxed{x=15-y} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Cmeasuredangle%20ACP%7D%7B%283x%2B2y%29%7D%2B%5Cstackrel%7B%5Cmeasuredangle%20BCP%7D%7B%283x%2B4y%29%7D%3D90%5Cimplies%206x%2B6y%3D90%5Cimplies%20%5Cstackrel%7B%5Ctextit%7Bcommon%20factor%7D%7D%7B6%28x%2By%29%7D%3D90%0A%5C%5C%5C%5C%5C%5C%0Ax%2By%3D%5Ccfrac%7B90%7D%7B6%7D%5Cimplies%20x%2By%3D15%5Cimplies%20%5Cboxed%7Bx%3D15-y%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A~%5Cdotfill)
![\bf \stackrel{\measuredangle APC}{(7x+3)}+\stackrel{\measuredangle BPC}{(16y)}=180\implies 7x+16y=177\implies 7\left( \boxed{15-y} \right)+16y=177 \\\\\\ 105-7y+16y=177\implies 9y=72\implies y=\cfrac{72}{9}\implies \blacktriangleright y=8 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \measuredangle BPC=16(8)\implies \measuredangle BPC=128](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Cmeasuredangle%20APC%7D%7B%287x%2B3%29%7D%2B%5Cstackrel%7B%5Cmeasuredangle%20BPC%7D%7B%2816y%29%7D%3D180%5Cimplies%207x%2B16y%3D177%5Cimplies%207%5Cleft%28%20%5Cboxed%7B15-y%7D%20%5Cright%29%2B16y%3D177%0A%5C%5C%5C%5C%5C%5C%0A105-7y%2B16y%3D177%5Cimplies%209y%3D72%5Cimplies%20y%3D%5Ccfrac%7B72%7D%7B9%7D%5Cimplies%20%5Cblacktriangleright%20y%3D8%20%5Cblacktriangleleft%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Cmeasuredangle%20BPC%3D16%288%29%5Cimplies%20%5Cmeasuredangle%20BPC%3D128)
Answer:
the answer would be 5b as the exterior angle is subdivided with the co efficient
Step-by-step explanation:
Answer:
The sum of the first 5 terms is -244
Step-by-step explanation:
To calculate the sum of the geometric series, we need the first term, the common ratio and the number of terms we would like to sum.
The first term here is -4
The common ratio is T2/T1 or T3/T2 = 12/-4 = -3
number of terms n = 5
The formula to use is;
Sn = a(1-r^n)/(1-r)
Plugging these values, we have;
Sn = -4(1-(-3)^5)/(1-(-3))
Sn = -4(1+243)/4 = -1(244) = -244
6 more
What was the start and end weight
Did they start at the same time maybe
