So...I'm stuck on this. HELP

Answers:

Yes the data can be modeled with a linear equation.

Processing fee = 42 dollars; Daily fee = 102 dollars

Max number of days you can rent house = 11 days

Explanation:

Part 1

Let

x = number of days

y = total cost in dollars

Note how each time x goes up by 2, y goes up by 204 (since 450-246 = 204). This consistent pattern directly tells us we have a linear equation going on. The next section goes into more detail.

Part 2

The daily fee is directly tied to the pattern mentioned earlier. We have the cost going up by $204 each time 2 days go by. So the daily fee is 204/2 = 102 dollars per day.

You could also use the slope formula with say the first two columns of the table

m = (y2-y1)/(x2-x1)

m = (450-246)/(4-2)

m = 204/2

m = 102

This confirms the daily fee is $102.

You can use any two columns of the table that you want to get the slope. It doesn't have to be the first two columns.

Now use (x,y) = (2,246) along with m = 102 to find the y intercept. Like before you can use any column you want to form the (x,y) value.

y = mx+b

246 = 102*2+b

246 = 204+b

246-204 = b

42 = b

b = 42

The equation is y = 102x+42. The initial processing fee is $42. If you stayed at the house for x = 0 days, you still have to pay the fee y = 42.

Part 3.

Plug in y = 1200 and solve for x

y = 102x+42

1200 = 102x+42

1200-42 = 102x

1158 = 102x

102x = 1158

x = 1158/102

x = 11.3529411764706

x = 11

We round down to the nearest whole number. This way we clear the hurdle.

Note that if x = 11, then,

y = 102x+42

y = 102*11+42

y = 1164

which is under the $1200 budget

In contrast, plugging in x = 12 leads to

y = 102x+42

y = 102*12+42

y = 1266

which is over the 1200 dollar budget. This shows that x = 11 is the max number of days you can rent the house.

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}

3.how is solving radical equations different than solving linear equations?

step-by-step explanation: