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2 years ago

You buy tomatoes at

Mathematics

1 answer:

Anna71 [15]2 years ago

6 0

Answer:

Expression:

1.20p + 3.99p

Now subsitutute p with the number of pounds given for both tomato and pepper.

1.20(5) + 3.99(2) = $13.98

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Suppose you invest $16000 at 9% interest and that it is compounded daily. How much will you have in 8 years?

kramer

Answer:

The amount after 8 years is $ 16,031.579

Step-by-step explanation:

Given as :

The Principal invested = $ 16000

The rate of interest compounded daily = 9 %

The time period = 8 years

Let The amount after 8 years = $ A

<u>From Compounded method </u>

Amount = Principal invested × $(1+\dfrac{\textrm Rate}{365\times 100})^{365\times \textrm Time}$

Or, Amount = 16000 × $(1+\dfrac{\textrm 9}{365\times 100})^{365\times \textrm 8}$

Or, Amount = 16000 × $(1.0002465)^{8}$

∴ Amount = $ 16,031.579

Hence The amount after 8 years is $ 16,031.579 Answer

4 0

3 years ago

If there are 16 red bobbles for every 2 green dandees, how many red bobbles are there in 20 green dandees?

kvasek [131]

Answer:

160

Step-by-step explanation:

20 / 2 = 10

10 * 16 = 160

8 0

2 years ago

In an 80/20 mortgage, what is the second mortgage used for?

avanturin [10]

Answer:

The first loan covers 80 percent of the home’s price, while the second covers the remaining 20 percent.

Step-by-step explanation:

they are still a home price percentage

5 0

3 years ago

Read 2 more answers

HI PLEASE HELP BELOW IM REALL IN NEED OF HELP

aalyn [17]

The answer is 3, because the other two sides are 3 and the angle is 60, so the other angles will have to be 60. That means it is an equilateral triangle.

3 0

3 years ago

What is the antiderivative of 3x/((x-1)^2)

Maslowich

Answer:

$\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

Step-by-step explanation:

Given

$\int \:\:3\cdot \frac{x}{\left(x-1\right)^2}dx$

$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$

$=3\cdot \int \frac{x}{\left(x-1\right)^2}dx$

$\mathrm{Apply\:u-substitution:}\:u=x-1$

$=3\cdot \int \frac{u+1}{u^2}du$

$\mathrm{Expand}\:\frac{u+1}{u^2}:\quad \frac{1}{u}+\frac{1}{u^2}$

$=3\cdot \int \frac{1}{u}+\frac{1}{u^2}du$

$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$

$=3\left(\int \frac{1}{u}du+\int \frac{1}{u^2}du\right)$

$\int \frac{1}{u}du=\ln \left|u\right|$ ∵ $\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)$

$\int \frac{1}{u^2}du=-\frac{1}{u}$ ∵ $\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1$

$=3\left(\ln \left|u\right|-\frac{1}{u}\right)$

$\mathrm{Substitute\:back}\:u=x-1$

$=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)$

$\mathrm{Add\:a\:constant\:to\:the\:solution}$

$=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

Therefore,

$\int \:3\cdot \frac{x}{\left(x-1\right)^2}dx=3\left(\ln \left|x-1\right|-\frac{1}{x-1}\right)+C$

4 0

3 years ago