Me! 100 points! ; ( question 10, 11, 12, 13, 14, 15. i have a f in this class and could use your a lot! i will love you forever if you me ; (

There are a lot of problems here so I'll try to be brief with each one so I don't add a lot of clutter

Problem 10) This is arithmetic because we subtract 6 from each term to get the next. This is the same as adding on -6. The common difference is d = -6. The first term is a = 1

Problem 11) Neither. The distance from 3 to 3/2 = 1.5 is 1.5 units, but the distance from 1.5 to 1 is 0.5 units. This implies there is no common difference value. The sequence is not arithmetic because of this. It's also not geometric either. To go from 3 to 3/2, we multiply by 1/2. But then going from 3/2 to 1, we multiply by 2/3. There is no common ratio r value.

Problem 12) This sequence is geometric. We divide each term by 3 to get the next, or put another way, we multiply each term by 1/3 to get the next term. The common ratio is r = 1/3. The first term is a = 108.

Problem 13) Neither. This is because going from term to term, we do not add the same amount each time. Example: from -2 to 4 we add 6, but then from 4 to -6 we add -10. So the sequence is not arithmetic. It's also not geometric either because we dont multiply by the same term each time. Eg: from -2 to 4, we multiply by -2; but from 4 to -6 we multiply by -1.5

Problem 14) The next three terms are: 1, 1/4, 1/16. This is found by multiplying each term by the common ratio r = 1/4, or you can think of dividing each term by 4. To get this common ratio, pick any term you want and divide it by its previous term. Example: term2/term1 = 16/64 = 1/4 = common ratio.

Problem 15) The next three terms are: -432, 2592, -15552. You multiply each term by the common ratio -6. Like with the previous problem, we divide any term over its previous one to get the common ratio, so for example, term2/term1 = -12/2 = -6 = common ratio.

mr. flanders’ class sold 47 doughnuts and   mr. rodriquez’s class sold 38 cartons of milk

step-by-step explanation:

let x be the number of doughnuts sold by mr. flanders’ class and y be the number of cartons sold by mr. rodriquez’s class.

1. together, the classes sold 85 items, then

x+y=85.

2. both classes earn earned $104.49 for their school. if doughnuts were sold for $1.35 each, then x doughnuts costed $1.35x. if cartons of milk were sold for $1.08 each, then y cartons costed $1.08y. hence,

1.35x+1.08y=104.49.

3. solve the system of two equations:

$4034.40

the perimeter of the field is the sum of the lengths of the four sides:

47 yd + 158 yd + 47 yd + 158 yd = 410 yd

the cost of 3 ft (1 yd) of fence is 3×$3.28 = $9.84.

so the cost of jake's wood perimeter fance is

(410 yd) × ($9.84 /yd) = $4034.40