prove that QR = QR by the reflexive property.
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
231 m2 is about 2,486.5 square feet.
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
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
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
sqr(e/pi)
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