Maria and edwin have $23.00 each to spend at a book fair, where all students receive a 25% discount. they both want to purchase a copy of the same book, which normally sells for $26.50 plus 10% sales tax. to check if she has enough to purchase the book, maria takes 25% of $26.50 and subtracts that amount from the normal price. she takes 10% of the discounted selling price and adds it back to find the purchase amount. edwin takes 75% of the normal purchase price and then computes 110% of the reduced price. is maria correct? is edwin correct? do they have enough money to purchase the book? explain your answer using complete sentences, and show your work.
@ogradyer

Step-by-step explanation:

They are both correct, the only difference is Maria did a two step process and Edwin did a one step process the total amount it would cost is $21.87 with $1.13 left over.

The way Maria calculates the final price of the book is correct. We know that the book normally sells for $26.50 plus a 10% sales tax. We also know that the students will receive an additional 25% discount. We must understand that the sales tax is applied to the discounted price. So first we will calculate 25% of 26.50 and subtract it from that amount. So the price will then be: (26.50 - 6.625). The price is now $18.34. Then to calculate the tax, it will be (10% * 18.34), so 1.834. We add this amount to the discounted price so: (18.34+1.834) = 20.12.

The final price of the book is $20.12 and Maria can get it and still save a little money.

your welcome

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

1: -2 3/4, -2 1/2, -14, 0, 3, 8 1/3

2: -4, -3, -2 1/2, -2 1/4, 3 1/4,

step-by-step explanation: