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3 years ago

The time you have before finance charges are assessed is called the a. credit limit. c. APR. b. incentive. d. grace period.

Mathematics

1 answer:

Tanya [424]3 years ago

6 0

Answer:

The time you have before finance charges are assessed is called the Grace period

option-D

Step-by-step explanation:

we know that

Grace period is the extra time given to customer to pay amount before finance charges

But once grace period passes , customer will have to pay extra fee or penalty with interest with passing days

So,

The time you have before finance charges are assessed is called the Grace period

So,

option-D

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Solve for x x−6/9=−5<br><br>x=

Vera_Pavlovna [14]

Answer:

4 4/9

Step-by-step explanation:

Move all terms not containing x

to the right side of the equation.

Exact Form:

x=−409

Decimal Form:x=−4.¯4

Mixed Number Form:

4 4/9

4 0

2 years ago

Read 2 more answers

Total world consumption of iron was approximately 900 million tons in 1993. If consumption had increased by 2.5% each year and t

Lilit [14]

Step-by-step explanation:

We can model the annual consumption of iron by the equation

c = 900 million tons x $1.025^n$ , where

n is the number of years after 1993 and

c is the consumption for that year.

An estimate for the total consumption of iron can now be obtained by integrating this equation with respect to a dummy variable, x from 0 to n.

$C = 900*10^6* \int\limits^n_0 {1.025^x} \, dx$

This equation simplifies to

C = 900 million tons x (1.025^n - 1) / ln(1.025), where

x is a dummy variable and

C is the total consumption of iron n years after 1993.

Let R be the amount of iron remaining at n years after 1993, so

R = 480 billion tons - C

R = 480 billion tons - 900 million tons x (1.025^n - 1) / ln(1.025)

We need to find the value of n when R is 0, so

0 = 480 billion tons - 900 million tons x (1.025^n - 1) / ln(1.025)

Simplify this equation to obtain

n = 107.4

So the iron will run out in the year 1993 + 107 = 2100

6 0

2 years ago

Find the Maclaurin series for f(x) using the definition of a Maclaurin series. [Assume that f has a power series expansion. Do n

aliya0001 [1]

Answer:

$f(x)=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}$

Step-by-step explanation:

The Maclaurin series of a function f(x) is the Taylor series of the function of the series around zero which is given by

$f(x)=f(0)+f^{\prime}(0)x+f^{\prime \prime}(0)\dfrac{x^2}{2!}+ ...+f^{(n)}(0)\dfrac{x^n}{n!}+...$

We first compute the n-th derivative of $f(x)=\ln(1+2x)$ , note that

$f^{\prime}(x)= 2 \cdot (1+2x)^{-1}\\f^{\prime \prime}(x)= 2^2\cdot (-1) \cdot (1+2x)^{-2}\\f^{\prime \prime}(x)= 2^3\cdot (-1)^2\cdot 2 \cdot (1+2x)^{-3}\\...\\\\f^{n}(x)= 2^n\cdot (-1)^{(n-1)}\cdot (n-1)! \cdot (1+2x)^{-n}\\$

Now, if we compute the n-th derivative at 0 we get

$f(0)=\ln(1+2\cdot 0)=\ln(1)=0\\\\f^{\prime}(0)=2 \cdot 1 =2\\\\f^{(2)}(0)=2^{2}\cdot(-1)\\\\f^{(3)}(0)=2^{3}\cdot (-1)^2\cdot 2\\\\...\\\\f^{(n)}(0)=2^n\cdot(-1)^{(n-1)}\cdot (n-1)!$

and so the Maclaurin series for f(x)=ln(1+2x) is given by

$f(x)=0+2x-2^2\dfrac{x^2}{2!}+2^3\cdot 2! \dfrac{x^3}{3!}+...+(-1)^{(n-1)}(n-1)!\cdot 2^n\dfrac{x^n}{n!}+...\\\\= 0 + 2x -2^2 \dfrac{x^2}{2!}+2^3\dfrac{x^3}{3!}+...+(-1)^{(n-1)}2^{n}\dfrac{x^n}{n}+...\\\\=\sum_{n=1}^{\infty}(-1)^{(n-1)}2^n\dfrac{x^n}{n}$