Can anyone explain both and help me with the answer as well? Thank you!

Suppose you have $13,000 to invest, which of the two rates would yield a larger amount in 1 year: 11% compounded daily or 10.86% compounded continuously
&
As x=-infinity, f(x)=?

11% compounded daily

Step-by-step explanation:

Simple interest is:

A = P (1 + r/n)^(nt)

where A is the final amount,

P is the initial amount,

r is the annual rate,

n is the number of compoundings per year (in this case, 365),

and t is the number of years.

A = 13,000 (1 + 0.11/365)^(365×1)

A = 14,511.37

Continuously compounded interest is:

A = Pe^(rt)

A = 13,000 e^(0.1086×1)

A = 14,491.31

11% compounded daily would yield a larger amount in 1 year.

Step-by-step explanation:

We would apply the formula for determining compound interest which is expressed as

A = P(1 + r/n)^nt

Where

A = total amount in the account at the end of t years

r represents the interest rate.

n represents the periodic interval at which it was compounded.

P represents the principal or initial amount deposited

From the information given,

P = $13,000

r = 11% = 11/100 = 0.11

n = 365 because it was compounded 365 times in a year.

t = 1 year

Therefore,

A = 13000(1 + 0.11/365)^365 × 1

A = 13000(1 + 0.0003)^365

A = 13000(1.0003)^365

A = $14504.1

The formula for continuously compounded interest is

A = P x e(r x t)

From the information given,

P = $13000

r = 10.86% = 10.86/100 = 0.1086

t = 10

year

Therefore,

A = 13000 x e(0.1086 x 1)

A = 13000 x e(0.1086)

A = $14491.3

the one that has a slope of 1/2 and is positive.

step-by-step explanation: 2 buckets 1 trip ; 4 buckets 2 trips and so on.

4.08%

step-by-step explanation: