In A Treatise on the Family (Cambridge, MA: Harvard University Press, 1981), Nobel laureate Gary Becker proposes his famous Rotten Kid Theorem as a sequential game between the potentially rotten child (player 1) and the child’s parent (player 2). The child moves first, choosing an action r that affects his own income !1ðrÞ½!01ðrÞ > 0+ and the income of the parent !2ðrÞ½!02ðrÞ < 0+. Later, the parent moves, leaving a monetary bequest L to the child. The child cares only for his own utility, U1ð!1 þ LÞ, but the parent maximizes U2ð!2 ! LÞ þ aU1, where a > 0 reflects the parent’s altruism toward the child. Prove that, in a subgame-perfect equilibrium, the child will opt for the value of r that maximizes !1 þ !2 even though he has no altruistic intentions. Hint: Apply backward induction to the parent’s problem first, which will give a first-order condition that implicitly determines L'; although an explicit solution for L' cannot be found, the derivative of L' with respect to r—required in the child’s first-stage optimization problem—can be found using the implicit function rule.

answer: domain and range mean all possible values of x and y that could be substituted to the inequality. a solution means all possible values that make the inequality statement true. so while the range is still all real numbers, the solution is the shaded area, for which y will always be greater than the line of the function.

answer: 7 x 2/5 = 2.8 hours, is your answer. your welcome! : 3

step-by-step explanation: first work out how long it took to hose 1/7 of the lawn which is 2/5 hours. times that by 7 to get the whole lawn.

sorry i don't know the other answer. i didn't learn what a proportional relationship is.

vertical asymptote .

horizontal asymptote .

step-by-step explanation:

given function is .

vertical asymptote is given by setting

denominator =0

when degree of numerator and denominator are same then divide first term of numerator by first term of denominator to get horizontal asymptote.

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.