A man stole $100 from a cashier counter in a shop when the cashier carelessly left check counter to the bathroom. Once the cashier returned, the thief used the money he stole to buy a pair of sneakers worth $70. How much money did the Cashier lose?
The thief stole $100 so the cashier has lost $100. Once the thief gives the $100 note to the cashier and buys the sneakers for $70, the cashier loses = $100-$70 = $30 The cashier then gives a $30 balance to thief resulting in a total loss of = 30 + 30 = $60 Another alternative to solving the problem : Think in terms of minus and plus The thief stole money from the cashier, so we have -$100 Once the thief gives buys sneakers from the cashier, we have a net of +$70 The cashier then gives a change of $30 back to the cashier give a net too of -$-30 Total net = -100 +70 -30 = -60 Thus -$60. (The negative sign implies loss)
Find the vertical and horizontal asymptotes of the function y= 5x/6x-3?
The asymptotes of a function refers to a particular value obtained when any part of a function is going to infinity. To solve this problem exactly we have to explore cases. (i) When x approaches +infinity (When x is going far far far right do we have a constant y value) (ii) When x approaches -infinity (When going far far far left do we have a constant y value) (iii) When y approaches +infinity (When going far far far up do we have a constant x value) (iv)When y approaches -infinity (When going far far far down do we have a constant x value) To get the horizontal asymptotes, simply look for the value for which the denominator is zero. Once you equate 6x-3 = 0. you get x=0.5 . This is the vertical asymptote. For the asymptote, find the limit of the function as x approaches infinity, To do this first simplify the expression by manipulation. Divide both the numerator and denominator by x and you'll have y= 5/6-3/x. 3/infinity is zero. So thus we have y= 5/6 for the horizontal asymptote. So thus the final answer is x=0.5 for the vertical asymptote and y=5/6 for the horizontal asymptote.
What is sin^-1(cos^2(x) x (1 + (tanx)^2))?
first solve what is in the bracket. 1 + (tanx)^2 is simply a substitution for sec^2 (x) Thus the question is simplified to sin ^-1 (cos^2(x) x sec^2(x)) since sec^2(x) is simply 1/cos^2(x), thus cos^2(x) x sec^2(x) = 1 thus the question simplifies to sin^-1 (1) which = 90 NB : Everything will work with you if you break the question into pieces.