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3 years ago

Two different examples of variables

1 answer:

4 0

Independent variables, which can be controlled or manipulated; dependent variables, which (we hope) are affected by our changes to the independent variables

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A $250 gaming system is discounted by 25%. It was then discounted another 10% because the box was opened by a customer. a. The s

iris [78.8K]

Answer:

a. Sale Price $=\$168.75$

b. Total percent discount $=32.5\%$

Step-by-step explanation:

First Discount:

Initial Price $=\$250$

Discount $=25\%$

$25\%\ of\ 250=\frac{25}{100}\times 250=62.5\\\\Now\ Price=250-62.5=$187.5$

Second Discount:

$Price=187.5\\\\Discount=10\%\\\\10\%\ of\ 187.5=\frac{10}{100}\times 187.5=18.75\\\\Now\ Sale\ Price=187.5-18.75=\$168.75$

Effective Discount:

Initial Price $=\$250$

Final Price $=\$168.75$

Total discount $=250-168.75=81.25$

$\%\ discount=\frac{81.25}{250}\times 100=32.5\%$

3 0

3 years ago

Solve the equation -3 = -13 - 5x for x

Law Incorporation [45]

Answer:

x = -2

Step-by-step explanation:

-3 = -13 - 5x

-3 + 13 = -5x

10 = -5x

10 ÷ (-5) = x

-2 = x

6 0

2 years ago

Read 2 more answers

The sum of two numbers is 9 . one number is nine less than twice the other. find the two numbers.

dimulka [17.4K]

Hello!

To solve this problem, we will use a system of equations. We will have one number be x and the other y. We will use substitutions to solve for each variable.

x+y=9
x=2y-9

To solve for the two numbers, we need to solve the top equation. The second equation shows that x=2y-9. In the first equation, we can replace 2y-9 for x and solve.

2y-9+y=9
3y-9=9
3y=18
y=6

We now know the value of y. Now we need to find x. We can plug in 6 for y in the second equation to find x.

x=2·6-9
x=12-9
x=3

Just to check, we will plug these two numbers into the first equation.

3+6=9
9=9

Our two numbers are three and six.

I hope this helps!

7 0

2 years ago

The total cost to attend the university Daniel wants to attend is going to be $12,000 for the first year.

Zielflug [23.3K]

Answer:

$395.83

Step-by-step explanation:

to solve, we first need to subtract what we already have, first the scholarship, wich is a set amount being taken from the original amount

12'000 - 2'500 = 9'500

now we have the second subtracting factor, but this one isn't set in stone and defined, it is half of the amount his parents will pay, so what we can do is divide what we have by two, wich will give us 2 halves

9'500 / 2 = 4'750

now all we have to do is divide again but this time for each month that Daniel needs to save up in, in this case 12

4'750 / 12 = 395.83 (note1)

and there we have it, that is the minimum amount Daniel would save each month

note1: (3 goes on for infinity, usualy this is represented by a line above the repeating number, this is the case of a repeating decimal)

5 0

3 years ago

Consider the following differential equation. x^2y' + xy = 3 (a) Show that every member of the family of functions y = (3ln(x) +

Veronika [31]

Answer:

Verified

$y(x) = \frac{3Ln(x) + 3}{x}$

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{x}$

Step-by-step explanation:

Question:-

- We are given the following non-homogeneous ODE as follows:

$x^2y' +xy = 3$

- A general solution to the above ODE is also given as:

$y = \frac{3Ln(x) + C }{x}$

- We are to prove that every member of the family of curves defined by the above given function ( y ) is indeed a solution to the given ODE.

Solution:-

- To determine the validity of the solution we will first compute the first derivative of the given function ( y ) as follows. Apply the quotient rule.

$y' = \frac{\frac{d}{dx}( 3Ln(x) + C ) . x - ( 3Ln(x) + C ) . \frac{d}{dx} (x) }{x^2} \\\\y' = \frac{\frac{3}{x}.x - ( 3Ln(x) + C ).(1)}{x^2} \\\\y' = - \frac{3Ln(x) + C - 3}{x^2}$

- Now we will plug in the evaluated first derivative ( y' ) and function ( y ) into the given ODE and prove that right hand side is equal to the left hand side of the equality as follows:

$-\frac{3Ln(x) + C - 3}{x^2}.x^2 + \frac{3Ln(x) + C}{x}.x = 3\\\\-3Ln(x) - C + 3 + 3Ln(x) + C= 3\\\\3 = 3$

- The equality holds true for all values of " C "; hence, the function ( y ) is the general solution to the given ODE.

- To determine the complete solution subjected to the initial conditions y (1) = 3. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y( 1 ) = \frac{3Ln(1) + C }{1} = 3\\\\0 + C = 3, C = 3$

- Therefore, the complete solution to the given ODE can be expressed as:

$y ( x ) = \frac{3Ln(x) + 3 }{x}$

- To determine the complete solution subjected to the initial conditions y (3) = 1. We would need the evaluate the value of constant ( C ) such that the solution ( y ) is satisfied as follows:

$y(3) = \frac{3Ln(3) + C}{3} = 1\\\\y(3) = 3Ln(3) + C = 3\\\\C = 3 - 3Ln(3)$

- Therefore, the complete solution to the given ODE can be expressed as:

$y(x) = \frac{3Ln(x) + 3 - 3Ln(3)}{y}$

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6 0

2 years ago