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3 years ago

Joe borrowed $7000 from the bank at a rate of 7% simple interest per year how much interest did he pay in six years

Mathematics

2 answers:

OlgaM077 [116]3 years ago

6 0

Answer:

Joe paid $2,940 in interest after six years.

Step-by-step explanation:

Since Joe borrowed $7000 from the bank at a rate of 7% interest per year. We need to find 7% of 7,000.

7,000 x 0.07 = 490

Since we need to find out how much interest Joe paid after six years, we need to multiply 490 by 6.

490 x 6 = 2,940

Therefore, Joe paid $2,940 in interest after six years.

Hope this helps! :D

tatiyna3 years ago

4 0

Answer:

840

Step-by-step explanation:

2,000(0.07) = 140

find the amount of interest per year

140(6) = 840

hope it helps!

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Answer:

Both length and width of 7.25m

Step-by-step explanation:

Let L and W be the length and width of the rectangle, respectively. Since the perimeter is 29, we have the following equation

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L = 14.5 - W

Also the area of the rectangle would be as the following:

A = LW = W(14.5 - W) = 14.5W - W^2

To find the maximum area possible, we can take the first derivative and set it to 0

A' = 14.5 - 2W = 0

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Determine whether the following are proportional or non proportional Y=1/2x

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Step-by-step explanation:

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3 years ago

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Lee las situaciones y realiza lo siguiente con cada una:

Julli [10]

Answer:

Part 1) see the explanation

Part 2) see the explanation

Part 3) see the explanation

Part 4) see the explanation

Step-by-step explanation:

The question in English is

Read the situations and do the following with each one:

Write down the magnitudes involved

Write which magnitude is the independent variable and which is the dependent variable

It represents the function that describes the situation

SITUATIONS:

1) A machine prints 840 pages every 30 minutes.

2) An elevator takes 6 seconds to go up two floors.

3) A company rents a car at S/ 480 for 12 days.

4) 10 kilograms of papaya cost S/ 35

Part 1) we have

A machine prints 840 pages every 30 minutes

Let

x ----> the time in minutes (represent the variable independent or input value)

y ---> the number of pages that the machine print (represent the dependent variable or output value)

Remember that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $k=\frac{y}{x}$ or $y=kx$

In this problem

we have a a proportional variation

The value of the constant of proportionality is equal to

$k=\frac{y}{x}$

we have

$y=840\ pages\\x=30\ minutes$

substitute

$k=\frac{840}{30}=28\ pages/minute$

The linear equation is

$y=28x$

Part 2) we have

An elevator takes 6 seconds to go up two floors.

Let

x ----> the time in seconds (represent the variable independent or input value)

y ---> the number of floors (represent the dependent variable or output value)

Remember that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $k=\frac{y}{x}$ or $y=kx$

In this problem

we have a a proportional variation

The value of the constant of proportionality is equal to

$k=\frac{y}{x}$

we have

$y=2\ floors\\x=6\ seconds$

substitute

$k=\frac{2}{6}=\frac{1}{3}\ floors/second$

The linear equation is

$y=\frac{1}{3}x$

Part 3) we have

A company rents a car at S/ 480 for 12 days.

Let

x ----> the number of days (represent the variable independent or input value)

y ---> the cost of rent a car (represent the dependent variable or output value)

Remember that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $k=\frac{y}{x}$ or $y=kx$

In this problem

we have a a proportional variation

The value of the constant of proportionality is equal to

$k=\frac{y}{x}$

we have

$y=\$480\\x=12\ days$

substitute

$k=\frac{480}{12}=\$40\ per\ day$

The linear equation is

$y=40x$

Part 4) we have

10 kilograms of papaya cost S/ 35

Let

x ----> the kilograms of papaya (represent the variable independent or input value)

y ---> the cost (represent the dependent variable or output value)

Remember that

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $k=\frac{y}{x}$ or $y=kx$

In this problem

we have a a proportional variation

The value of the constant of proportionality is equal to

$k=\frac{y}{x}$

we have

$y=\$35\\x=10\ kg$

substitute

$k=\frac{35}{10}=\$3.5\ per\ kg$

The linear equation is

$y=3.5x$

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