Find the derivative of $$y=sin(3x^2+2x+1)$$.
We will use the chain rule to solve for this derivative. To use the chain rule we will first take the derivative of the outside term, in this case, $$sin$$. Then we multiply it by the derivative of the inside term, in this case, $$3x^2+2x+1$$ Applying this we have $$y' = cos(3x^2+2x+1) (6x+2)$$.
There are 4 green beads, 6 red beads, and 10 blue beads in a jar. Assuming the beads are drawn at random, what is the probability that 2 red beads will be selected in a row without replacement?
Let's start with the probability that one red bead is selected. Which would be calculated as, $$6/(4+6+10) = 6/20$$ or $$3/10$$. Now we will incorporate the second red bead. The problem indicates that the first red bead drawn would not be replaced. This now means there are now only 5 red beads and only 19 total beads left in the jar. So the probability of drawing another red bead would be $$5/19$$ To calculate the overall probability that these two events happen in a row, we will multiply their independent probabilities. So we have $$(3/10)(5/19)=15/190$$. Simplifying we have $$3/38$$,
Solve the system of linear equations: $$(1) 4x + 5y = 29$$ $$(2) 2x + y = 10 $$
We will use elimination to solve this system of equations. First multiply equation (2) by 2, $$2(2x + y = 10)$$ $$= 4x+2y=20.$$ Then subtract our new form of equation (2) from equation (1) and we have $$ 4x + 5y = 29$$ $$-(4x+2y =20)$$ Combining like terms, the $$x$$s cancel and we are left with $$3y=9$$. $$y=3$$ We have now solved for one of the two variables. Now substitute $$y=3$$ back into either of the original equations to solve for $$x$$. Substituting into equation (2) we have $$2x+3=10$$ $$2x=7$$ $$x=7/2$$ or $$3.5$$