Area is not measured in centimeter units. It may be guessed that the question should have indicated the area in square centimeters.The area of a regular pentagon is given as:A = s * s * sqr(25 + 10*(sqr(5)) / 4Solve for s:s = 2 * sqr (A / sqr(25 + 10*(sqr(5)))Insert the area of 600 square centimters:s = 2 * sqr (600 cm*cm / sqr(25 + 10*(sqr(5)))s = aproximately 19.67 cm
392040 sqr ft
I'm assuming that these are 3-D coordinates A(0,0,0) and B(3,4,5) The distance formula between two points in 3 dimensions is the following: D = SQR {(x1-x2)2 + (y1-y2)2 + (z1-z2)2} D = SQR {(0-3)2 + (0-4)2 + (0-5)2} D = SQR {9+16+25} = SQR {50} D is approx. = 7.07
pythagoreon thereom 34sq - 30sq = 256 sqrt = 16 base 16+16 = 32 half length of base = SQR(34^2 - 30^2) =SQR(1156-900) =SQR(256) =16 base length = 2x 16=32
Pi x D2 / 4 = area ft2 = 3.142*8*8/4 = 50.272 square yard = 9 sqr ft (3*x3) thus 50.272/9 sqr yrds in 8 ft diameter = 5.586
Bisector of an angle in basically a line which is drawn from the vertex of the angle and bisect's or cuts the angle into 2 halves. For example we have angle PQR and if we cut a bisector through it then like: QS then SQR = 1/2*PQR
The sin of 30 is sqr(3)/2 so 6 * sqr(30) is 6 * sqr(3)/2 this is also 6/2 * sqr(3) which simplifies to 3 * sqr(3) which is approximately 5.19615242
Area is not measured in centimeter units. It may be guessed that the question should have indicated the area in square centimeters.The area of a regular pentagon is given as:A = s * s * sqr(25 + 10*(sqr(5)) / 4Solve for s:s = 2 * sqr (A / sqr(25 + 10*(sqr(5)))Insert the area of 600 square centimters:s = 2 * sqr (600 cm*cm / sqr(25 + 10*(sqr(5)))s = aproximately 19.67 cm
231 m2 is about 2,486.5 square feet.
sqr(-30) is i*sqr(30) We use i to represent the sqr(-1) because it is called an imaginary number. It takes an imagination to work with it. It is not fictitious.
x=81
110
x=81
54 = 625 (to find out sqr rt 625 then sqr rt again)
main() { sum=0; float avg=0.0; int sqr[10]; for(i=1;i<=10;i++) { sqr[i]=(i*i); } for(i=1;i<=10;i++) { sum=sum+sqr[i]; } avg=sum/10; dts it !!!
8(1st eight) +8(2nd eight)= 16-> sqr(16)=4 -> sqr(4)=2 ; 8(3rd eight)-2=6
2x2-2x=12x2-2x-1=0a=2 b=-2 c=-1x= (-b +- sqr(b2-4ac))/2ax= (2 +- sqr(4 + 8))/4x= (2 +- sqr(12))/4x = 1/2 +- sqr(12)/4x ~= .5 +- 0.8660x ~=-0.3660 or x ~= 1.3660