diboron hexahydride
diboron hexahydride
A=1/2h(a+b)
P= 2l+2w
This is a good question, and you need to know the quadratic formula let A = area b = width h = height P = perimeter A = bh b = A/h P = 2b + 2h P = 2A/h +2h = (2h^2 + 2A)/h Ph = 2A + 2h^2 2h^2 - Ph + 2A = 0 use quadratic formula to solve: h =( P +/- SQRT(P^2 -16A))/4 This gives both sides b and h using plus then minus the square root
Another name for is is
Suppose base and height are b and h. 2b+2h=44 (the perimeter includes two bases and two widths) b=2b-11 base is 11 less than two times base Subtract b from both sides: 0=2b-11-b, b-11=0, b=11 2b=22 2b+2h=44, 22+2h=44, 2h=44-22=22, h=11 So the rectangle is a square with side 11.
Perimeter of a rectangle (P) = 2b+2h = 2*24+2h=114 48+2h=114 2h=66 h=33" Area of a rectangle (A) = b*h = 24*33= 1584 sq. in.
perimeter = 2h + 2b ....where h = height and b = base you say that the h = (2b+3) so 2h=2x(2b+3) so substitute that in and we have: perimeter = (2 x (2b +3)) + (2 x b) multiplying out gives perimeter = 4b + 6 + 2b = 6b+6 now we know the perimeter = 72mm = 6b+6 taking the 6 off both sides we get 6b=66mm divide by 6 and we find b=11mm we know h = 2b + 3 and that b=11 so h=22+3=25mm now Area of a rectangle = b x h = 11mm x 25mm = 275mm^2
A=1/2h(b1+b2) where h is the height and b is the base
B = 2H - 3; (B x H)/2 = 10 ie B x H = 20;2H2 - 3H - 20 = 0(H - 4)(2H + 5) = 0Only positive value for H is 4, making B = 5 cm
If the pentagon is a regular pentagon, then each of the central angles (formed by joinning the center with the vertices) is72⁰ (1/5 of 360⁰). So that 5 congruent isosceles triangle are formed, with base angles of 54⁰. If we draw the height, of these triangles (from the center to the sides of the pentagon), 10 congruent right triangles are formed. Let denote with: b the sides of the pentagon, and h the height. In one of the right triangles we have: tan 54⁰ = h/(b/2) tan 54⁰ = 2h/b 1.38 = 2h/b 2h = 1.38b h = (1.38/2)b h = .69b So that, h/b = .69b/b = .69
The binary compound formed from the elements with the symbols B and O is boron oxide, with the chemical formula B2O3.