Let σ∈sx. if σn(x)=y, we will say that x∼y. i. show that ∼ is an equivalence relation on x. ii. if σ∈an and τ∈sn, show that τ−1στ∈an.τ−1iii. define the orbit of x∈x under σ∈sx to be the setox,σ={y: x∼y}.compute the orbits of each of the following elements in s5: α=(1254)β=(123)(45)γ=(13)(25).iv. if ox,σ∩oy,σ≠∅,ox, prove that ox,σ=oy,σ.the orbits under a permutation σ are the equivalence classes corresponding to the equivalence relation ∼.v. a subgroup h of sx is transitive if for every x, y∈x, there exists a σ∈h such that σ(x)=y. prove that 〈σ〉 is transitive if and only if ox,σ=x for some x∈x(abstract algebra)

let's cover over the formula of the vertex:

y = a(x – h)^2 + k

in this formula, (h,k) are the vertex, always remember that the positive or negative value of the vertex is based on the sign it has.

for example, if the vertex is -3,-6 then both the x and y value would have the positive sign attached to it and if it were a positive x and y value, then a negative sign would be attached to it.

with that being said, for our problem the x value is a positive one so we are looking for a number with a negative sign next to it, in this case the first and second choice.

however, for the first answer choice it isn't quite part of the formula so we know that the answer choice correctly displays the x value.

to confirm, we can see if the -1 fits correctly with the second answer choice, and as mentioned above a negative number will have a positive sign attached to it. so, for our problem we can see that -1 has been changed to +1, so we can confirm that the second answer choice is correct.

hence, our final answer is y=(x - 1)^2 + 1

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.

what are your options?

step-by-step explanation: