A right triangle has a short leg equal to 6 and a long leg equal to 8. Find the hypotenuse using the Pythagorean Theorem.
The Pythagorean Theorem is a^(2) + b^(2) = c^(2), where a= short leg, b= long leg, and c= hypotenuse. So we have a=6, b=8, and must solve for c. Plug the values in the Pythagorean Theorem equation (6)^2 + (8)^2 = c^2 36 + 64 = c^2 100 = c^2 10 = c Therefore our hypotenuse equals 10. Our right triangle has sides equal to 6, 8, and 10 (from shortest to longest leg).
Find the indefinite integral of x^(2)−3x−4.
Because the integral is indefinite, we only need to solve the problem as is, and not worry about plugging in values after integrating. x^(2)-3x-4 Add one to each exponent value, and place that same number as a denominator. x^(2+1)/3 - 3x^(1+1)/2 - 4x^(0+1)/1 [x^(3)]/3 - [3x^(2)]/2 -4x Check you answer by deriving. If you get the same answer as the initial problem, then you solved correctly. When you derive, you move the exponent to the front and subtract 1 from the exponent of each variable. [3*1x^(3-1)]/3 - [2*3x^(2-1)]/2 - [1*4x^(1-1)] [3x^(2)]/3 - [6x^(1)]/2 - [4x^(0)] x^(2) - 3x - 4 Remember, anything to the power of 0, equals 1. Since we derived and got the same problem as we started, our integration was done correctly.
Solve the system of equations using the elimination method: 5x+3y=5 10x-3y=25
Under the elimination method our goal is to cancel one variable initially, or in the occasion that both variables cancel out, there is no solution. 5x+3y=5 5x-3y=25 Multiplying a variable to cancel with another is not necessary in this example because the y's already cancel (3y-3y=0). 5x=5 5x=25 Next step is to add straight down. Make sure to add the correct variables. No mismatching. 10x=30 Solve. Divide both sides by 10 to solve for x. x=3 Now that we have solved for x, we can plug in our answer in to either equation to solve for y. Let's plug x=3 into the first equation. 5(3)+3y=5 15+3y=5 3y= -10 y= -10/3 Finally we get the two values for x and y. x=3, y= -10/3