Jamesaiden

105=3*5*7

total : 3!=3*2*1=6

perimeter = \(2r+\frac{1}{4}\times\pi \times 2r=72+18\pi\)

x = 180 - 43 - (180 - 126)

x = 126 - 43 = 83

\(\frac{1}{997}\equiv x\pmod {1000}\)

997|1000(1000-a)+1

(1000-3)|(1000)(1000-a)+1

last 3 digits from (1000)1000-a+1 = 001.

then, last 3 digits from 1000-3 is 001 too.

chance : (1000-3)*333.

=> 997*333.

x=333

2.) Area : \(\frac{1}{2}\times AC\times BD\)

\(= \frac{1}{2}\times 4\sqrt{5}\times \sqrt{(5\sqrt{2})^2-(3\sqrt{5})^2}\)

\(= \frac{1}{2}\times 4\sqrt{5}\times \sqrt{5}=10\)

1.) Area of the rectangle :

\(w=4\times r=4\times 3=12 \)

\(h=r+\sqrt{(2r)^2-r^2}+r=2\times 3+3\sqrt{3}=6+3\sqrt{3}\)

Area : \(w\times h=12\times (6+3\sqrt{3})=72+36\sqrt{3}\)

Hope this help ^_^

\(AC=AD=26\)

\(CG=GD=24\)

\(AG=\sqrt{CA^2-CG^2}=\sqrt{676-576}=10\) (H)

(2x - 5) + (2x + 5) = 180

4x = 180

x = 45.

\(\angle P=x+10=55\)

\(\angle S=2x+5=95\)

\(\angle Q=2x-5=85\)

\(\angle R=180-\angle P=125\)

Thanks

\(MQ=QR=RP\)

\(\angle QMR=\angle QRM=x\)

\(\angle MQR=180-2x\)

\(\angle RQP=\angle RPQ=180-(180-2x)=2x\)

\(\angle QRP=180-2x- 2x=180-4x\)

\(\angle MRN=180-(180-4x)-x=3x\)

because \(\angle NMR=\angle NRM \) we know that \(MN=NR\)

\(MN=NQ=NR\)

\(\angle NMQ=\angle NQM=\angle NRQ=\angle NQR=4x\)

\(\angle MQN+\angle NQR+\angle RQP=180\)

\(4x+4x+2x=180\)

\(x=18\) (C)