Find two numbers whose sum is 26 and whose product is 165

Let the two numbers be x and y Given: x + y = 26 -------eq(1) Given: x*y = 165-------eq(2) From eq(2), making y the subject of the formula, y = 165/x Substitute this value into eq(1) Eq(1) becomes, x + (165/x) = 26 Now we solve for x in the above equation. multiply both sides by x, x^2 + 165 = 26x ==> x^2 - 26x +165 = 0 This is a quadratic equation solving this quadratic equation, we have x = 11 or 15 substituting this value of x into eq(1) and solving for y y= 15 or 11. Therefore the two numbers are 11 and 15

Find the derivative of y with respect to x for the function: y = 2x^2 - 5x + 1

Dy/Dx of aX^n = anX^n-1 General formula for the derivative of y with respect to x Note the Dy/Dx of a Constant is 0 Therefore Dy/Dx of 2x^2 - 5x + 1 is..... 2*2*x^(2-1) -5*1*x^(1-1) + 0 = 4x-5

It took Alfred 3.5 hours to drive from city A to city B. On his way back to city A, he increased his speed by 20 km per hour and it took him 3 hours. What is the average speed of the whole journey?

First, the formula for calculating Speed is Distance/ Time Average Speed = Total Distance / Total Time This implies that Distance= Speed * Time Let the speed at which Alfred traveled from A to B be x km/h Time(A to B) = 3.5hrs This means that Alfred drove at a speed of (x+20) km/h on his way back i.e. From B to A Time(B to A) = 3hrs => Average Speed = [3.5x + 3(x+20)] / 3 + 3.5 Km/h Note Distance From A to B is the same as the Distance from B to A. Therefore, 3.5 * x = 3 (x+20) Expanding the bracket, 3.5x = 3x + 60 Solving for x, 3.5x-3x = 60 ===> x = 60/ 0.5 = 120km/h This implies that Alfred Traveled from A to B at 120km/h in 3.5hrs, and from B to A at 140km/h in 3hrs Average Speed = [3.5*120 + 3(120+20)] / 3+ 3.5 = 129.2km/h