The area of a right triangle is 50. One of its angles is 45o. Find the lengths of the sides and hypotenuse of the triangle.
The triangle is right and the size one of its angles is 45 degree ; the third angle has a size 45 degree and therefore the triangle is right and isosceles. Let x be the length of one of the sides and H be the length of the hypotenuse. Area = (1/2) x Square = 50 , solve for x: x = 10 We now use Pythagoras theorem to find H: x square + x square = H square Solve for H: H = 10 sqrt(2)
If collector current Ib is 100uA and Beta = 100 .Calculate Ie
We Know Ie = Ic+Ib and Ic = Beta * Ib hence Ie = ( Beta + 1 ) * Ib = 101 * 100 u = 10.1 mA
If x <2, simplify |x - 2| - 4|-6|
Given the expression |x - 2| - 4|-6| STEP 1. If x < ;2 then x - 2 < 2 and if x - 2 < 2 the |x - 2| = -(x - 2). STEP 2. Substitute |x - 2| by -(x - 2) and |-6| by 6 . |x - 2| - 4|-6| = -(x - 2) -4(6) = -x -22