20 POINTS!!

When you study algebra, you see relationships in the form of graphs, tables, and equations. These relationships connect independ

Llana [10]

Take an example: The salary per hour:

Assume x (variable) is the number of hours worked and assume THAT THE RATE PER HOUR IS $10, then:
If you work x hours at the hourly rate of$10, you will generate an income,
say, y that is equal the number of hours worked by the rate per hour, that is:

y = 10.x. The more hour(x) you work, the more income(y) you get.
Hence we created a relationship between y and x , or in English, between the earning and the number of hours worked.
Moreover y depends on x, if there is no x, there is no y and if x increases, y increases. However x is an INDEPENDENT variable because you can
give to x any number & y will follow

3 0

2 years ago

!!!!!!!!!!!!!!!!!!!!

Andru [333]

Answer:

$f^{-1}(x)=4x+20$

Step-by-step explanation:

Given

$f(x)=\frac{x}{4}-5\\ \\y=\frac{x}{4}-5$

Swap x and y and solve for y

$y=\frac{x}{4}-5\\ \\x=\frac{y}{4}-5\\ \\x+5=\frac{y}{4}\\ \\4(x+5)=y\\\\y=4x+20$

Final answer

$f^{-1}(x)=4x+20$

Keep in mind that we need to swap the variables because a function and its inverse are symmetrical about the line y=x. See the graph below.

3 0

2 years ago

Use undetermined coefficient to determine the solution of:y"-3y'+2y=2x+ex+2xex+4e3x

Kitty [74]

First check the characteristic solution: the characteristic equation for this DE is

r ² - 3r + 2 = (r - 2) (r - 1) = 0

with roots r = 2 and r = 1, so the characteristic solution is

y (char.) = C₁ exp(2x) + C₂ exp(x)

For the ansatz particular solution, we might first try

y (part.) = (ax + b) + (cx + d) exp(x) + e exp(3x)

where ax + b corresponds to the 2x term on the right side, (cx + d) exp(x) corresponds to (1 + 2x) exp(x), and e exp(3x) corresponds to 4 exp(3x).

However, exp(x) is already accounted for in the characteristic solution, we multiply the second group by x :

y (part.) = (ax + b) + (cx ² + dx) exp(x) + e exp(3x)

Now take the derivatives of y (part.), substitute them into the DE, and solve for the coefficients.

y' (part.) = a + (2cx + d) exp(x) + (cx ² + dx) exp(x) + 3e exp(3x)

… = a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)

y'' (part.) = (2cx + 2c + d) exp(x) + (cx ² + (2c + d)x + d) exp(x) + 9e exp(3x)

… = (cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

Substituting every relevant expression and simplifying reduces the equation to

(cx ² + (4c + d)x + 2c + 2d) exp(x) + 9e exp(3x)

… - 3 [a + (cx ² + (2c + d)x + d) exp(x) + 3e exp(3x)]

… +2 [(ax + b) + (cx ² + dx) exp(x) + e exp(3x)]

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

… … …

2ax - 3a + 2b + (-2cx + 2c - d) exp(x) + 2e exp(3x)

= 2x + (1 + 2x) exp(x) + 4 exp(3x)

Then, equating coefficients of corresponding terms on both sides, we have the system of equations,

x : 2a = 2

1 : -3a + 2b = 0

exp(x) : 2c - d = 1

x exp(x) : -2c = 2

exp(3x) : 2e = 4

Solving the system gives

a = 1, b = 3/2, c = -1, d = -3, e = 2

Then the general solution to the DE is

y(x) = C₁ exp(2x) + C₂ exp(x) + x + 3/2 - (x ² + 3x) exp(x) + 2 exp(3x)

4 0

2 years ago

How would we solve 6/15=2/c

eduard

The answer is c=5 hope this helps

5 0

3 years ago

Please Help! Need by today!

ra1l [238]

Answer:

sorry but no one answered it sorry good lucked

Step-by-step explanation:

4 0

1 year ago

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