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Black_prince [1.1K]
3 years ago
14

Sarah invested $800 in an account paying an interest rate of 3.5% compounded quarterly. Assuming no deposits or withdrawals are

made, how much money, to the nearest cent, would be in the account after 8 years?

Mathematics
1 answer:
Korolek [52]3 years ago
5 0

Answer:

$ 1,057.22

Step-by-step explanation:

A = $ 1,057.22

A = P + I where

P (principal) = $ 800.00

I (interest) = $ 257.22

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VARVARA [1.3K]

<span>Je ne suis pas sûr de ce que vous voulez dire?</span>
7 0
3 years ago
Please help if you can!!!!
storchak [24]

Answer:Well, I don't know what you got so I can't tell you if it is right.

If it works in both equations, it depends of whether your equations are set up correctly.

Here is how I would do this problem.

Let x = no. of hot dogs,y = number of sodas.

First equation is just about the number of things.

x + y = 15

Second equation is about the cost of things.

1.5 x + .75 y = 18

solve x+y = 15 for y  y = 15-x    substitute into second equation

1.5x + .75(15 - x) = 18    

You should get the correct answer for number of hot dogs if you solve this correctly.  Put your answer in the x + y =15 equation to get y.  Then put both x and y into the cost equation and check your answer.

Hope this helps.

Step-by-step explanation:

5 0
2 years ago
Hello can someone please help me with this.
Zolol [24]

Answer:

if there is equivalent ratio cookies should be 48

3 0
3 years ago
I'LL WON'T BE OFFEND IF U DON'T HELP ME WITH THIS
Evgesh-ka [11]

The average is found by adding the numbers and dividing by 2.

Since the average is 17, the total of the two numbers must equal 17 x 2 = 34.


One number is given as 11, so subtract 11 from 34:

34 - 11 = 23

The other number is 23.

Too check: add 11 and 23 and then divide by 2:

11 + 23 = 34

34/2 = 17.

5 0
3 years ago
Read 2 more answers
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}

3 0
2 years ago
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