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3 years ago

Find the simple interest earned

Mathematics

1 answer:

Kitty [74]3 years ago

6 0

Answer:

SI = $1,200 ; A = $6,200

Step-by-step explanation:

First, converting R percent to r a decimal

r = R/100 = 3%/100 = 0.03 per year,

then, solving our equation

I = 5000 × 0.03 × 8 = 1200

I = $ 1,200.00

The simple interest accumulated

on a principal of $ 5,000.00

at a rate of 3% per year

for 8 years is $ 1,200.00.

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JulsSmile [24]

Answer: its b ;p

Step-by-step explanation:

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3 years ago

Read 2 more answers

Question 1

Ilia_Sergeevich [38]

We will see that the perimeter of the rectangle is exactly 390 ft, so the statement is true.

<h3>Is the fence enough?</h3>

For a rectangle of length L and width W, the perimeter is given by:

P = 2*(L + W).

In this case, we know that:

L = 120 ft

W = 75 ft

Replacing that on the perimeter equation we get:

P = 2*(120 ft + 75 ft) = 390 ft

So the perimeter is exactly 390 ft, meaning that to put a fence around the parking lot the company will need at least 390 ft of fence.

So the statement is correct.

If you want to learn more about perimeters, you can read:

brainly.com/question/24571594

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2 years ago

An advertising company designs a campaign to introduce a new product to a metropolitan area of population 3 Million people. Let

Advocard [28]

Answer:

$P(t)=3,000,000-3,000,000e^{0.0138t}$

Step-by-step explanation:

Since P(t) increases at a rate proportional to the number of people still unaware of the product, we have

$P'(t)=K(3,000,000-P(t))$

Since no one was aware of the product at the beginning of the campaign and 50% of the people were aware of the product after 50 days of advertising

We have and ordinary differential equation of first order that we can write

$P'(t)+KP(t)= 3,000,000K$

The <em>integrating factor </em>is

$e^{Kt}$

Multiplying both sides of the equation by the integrating factor

$e^{Kt}P'(t)+e^{Kt}KP(t)= e^{Kt}3,000,000*K$

Hence

$(e^{Kt}P(t))'=3,000,000Ke^{Kt}$

Integrating both sides

$e^{Kt}P(t)=3,000,000K \int e^{Kt}dt +C$

$e^{Kt}P(t)=3,000,000K(\frac{e^{Kt}}{K})+C$

$P(t)=3,000,000+Ce^{-Kt}$

But P(0) = 0, so C = -3,000,000

and P(50) = 1,500,000

$e^{-50K}=\frac{1}{2}\Rightarrow K=-\frac{log(0.5)}{50}=0.0138$

And the equation that models the number of people (in millions) who become aware of the product by time t is

$P(t)=3,000,000-3,000,000e^{0.0138t}$

5 0

3 years ago

What is the solutions to 0=x^2-x-6<br> two things <br> ,,quadratic equationz,,,

natita [175]

Answer:

x1=3. x2=-2

Step-by-step explanation:

$x = \frac{ - b + - \sqrt{b {}^{2} - 4ac } }{2a}$

b=-1

a=1

c=-6

$x = \frac{ - ( - 1) + - \sqrt{ {( - 1)}^{2} - 4 \times 1 \times ( - 6) } }{2 \times 1}$

$x = \frac{1 + - \sqrt{1 + 24} }{2}$

$x = \frac{1 + - \sqrt{25} }{2}$

$x1 = \frac{1 + 5}{2}$

$x2 = \frac{1 - 5}{2}$

x1=3

x2=-2

$x = \frac{1 + - 5}{2}$

3 0

3 years ago

Please help me answer this!!

tangare [24]

Answer:

Here,

1-4

2-3

3-1

4-2

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3 years ago