AboutAbe Mantell Expertise Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!
Experience Over 15 years teaching at the college level.
3. Find the height, to the nearest tenth of a foot, of the tower showwn. Both sides of the triangle is 50ft and the base is 10ft wide. ( it looks like an isosceles triangle)
4. Simplify : (3+5sqroot2)(8-sqroot2)
Sorry for bombarding with questions. I am stuck on these, do you mind showing me the steps!
Much appreciated
Answer Hello Anne,
1. I gather you mean the cube-root, yes? cube_root(81a^6b^8), factor out perfect cubes...
. cube_root(27*3*a^6*b^6*b^2), now the 27a^6b^6 can came out of the cube_root..as:
. 3a^2b^2*cube_root(3b^2)
2. sqrt(100m^5n)/(sqrt(5m^3)=sqrt(100*m^4*m*n)/sqrt(5m^2*m)=10m^2*sqrt(m*n)/(m*sqrt(5m))
. which can be simplified to: 10m^2*sqrt(n)/(m*sqrt(5))
3. If it is an isosceles triangle, then form a right triangle by drawing the height from
. the top to the base...then you have (two) right triangles with height h, base 5, and
. hypotenuse 50...now use the Pythagorean Thrm: h^2+5^2=50^2, so h^2=2475
. h=sqrt(2475)=15*sqrt(11) or about 49.75 ft
4. (3+5sqrt(2))(8-sqrt(2)), multiply by the conjugate of the denominator (8+sqrt(2))/(8+sqrt(2))
. to get: (14+37*sqrt(2))/60
