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3 years ago

If something cost2.99$ what is the total cost with taxes

1 answer:

6 0

Without a lot more information, nobody can say. Every city, state, and county
can have its own taxes. They may be at a different rate from other cities, states,
and counties, and they may apply to different items.

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V5. Suppose you invest $5,000 at 9% interest, compounded annually, for 10 years. Determine the future value of your investment,

Elanso [62]

Answer: $11836.8

Step-by-step explanation:

Given. That :

Amount invested = $5000

Interest rate = 9% = 0.09

Period = 10 years, compounded annually

Using the compound interest formula :

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

A = final amount

P = principal or invested amount

r = rate of interest

n = number of times interest Is applied per period

t = period

A = 5000(1 + 0.09/1)^(1*10)

A = 5000(1.09)^10

A = 5000 * 2.36736367459211723401

A = 11836.81837296058617005

= $11836.8

4 0

3 years ago

Help Ill give 100 points!

nata0808 [166]

Answer:

0, 1, 1/2

Step-by-step explanation:

First you would have to make them have the same denominators.

The lowest one they share is 24.

What ever you do to the bottom you have to do to the bottom.

1/6, times each by 4.

1 * 4 = 4

6 * 4 = 24

4/24.

Now you have to do the same thnig for 5/8.

But times it by 3, since 8 times 3 is 24.

5 * 3 = 15

8 * 3 = 24

15/24

Now subtract.

15 - 4 = 11/24

the difference is close to one half.

4/24 = 1/6 is close to 0.

15/24 = 5/8 is close to 1.

4 0

3 years ago

Find the GCF of 4y and 12y

love history [14]

Answer:

Step-by-step explanation:

The factors of 4 are: 1, 2, 4

The factors of 12 are: 1, 2, 3, 4, 6, 12

Then the greatest common factor is 4.

Hope this helps! :D

Please mark brainlist

7 0

3 years ago

The scale factor between two similar shapes is 4. what is the scale factor of area

lana [24]

I think the scale of the factor of area is 2

6 0

3 years ago

Read 2 more answers

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

4 0

3 years ago