It is: 9h+4(2+h) = 9h+8+4h => 13h+8
Its height (h) is side (s) times sin 60 orsqrt(3)/2 times side = .866s h = .866s s = h/.866 = 2/sqrt(3) s = 1.154 h perimeter = s + s + s = 3s = 3 x 2/sqrt(3) = 3.46s
Which of the following formulas is used to find the area of a trapezoid? Solution: 1/2 h(B+b) The area of a trapezoid is 1/2 times the height times the sum of both bases. h is the height, b is the top base and B is the bottom base. A trapezoid=1/2×h×(B+b)
This is a nice bit of calculus. You sum up an infinite number of infinitely thin slices of the cone as follows: Radius of each slice = r*(h-x)/h Surface area of each slice = Pi*r2(h-x)2/h2 Integral of Pi*r2(h-x)2/h2 dx evaluated from 0 to h = Pi*r2*h/3
15
It is: 9h+4(2+h) = 9h+8+4h => 13h+8
6h
It means that you multiply 2 by any given number (h) and then multiply by 5. h is the variable. A variable is anunknown number. So if h is 3, then you replace it with h so 2 times 3 is 6 times 5 is 30.
Its height (h) is side (s) times sin 60 orsqrt(3)/2 times side = .866s h = .866s s = h/.866 = 2/sqrt(3) s = 1.154 h perimeter = s + s + s = 3s = 3 x 2/sqrt(3) = 3.46s
in singles match between each other 4-3 to triple h and in tag team matches 2-2 or tie
Which of the following formulas is used to find the area of a trapezoid? Solution: 1/2 h(B+b) The area of a trapezoid is 1/2 times the height times the sum of both bases. h is the height, b is the top base and B is the bottom base. A trapezoid=1/2×h×(B+b)
This is a nice bit of calculus. You sum up an infinite number of infinitely thin slices of the cone as follows: Radius of each slice = r*(h-x)/h Surface area of each slice = Pi*r2(h-x)2/h2 Integral of Pi*r2(h-x)2/h2 dx evaluated from 0 to h = Pi*r2*h/3
The expression n2 + h2 + nh3 is the sum of the squares of two numbers n^2 and h^2, along with the product of n and h multiplied by 3.
15
3/10*h or 0.3*h
3+h
-1