Algebra Help Functions
Can you help my solve these math problems? Algebra, Functions and Equations?
Please can you tell me how to solve it also?
Algebra
x squared multiply by x cube divided by x tot he power of four
a3/2b5/2 multiply by the square root of ab cube (the 3/2 is above the a and the 5/2 is above the b
Functions
f(x) = 2x-3
What is f inverse(0)? What is f inverse f(2)?
Simultaneouss Equations
How do you solve the equations for x and y
y+4x=27
xy+x=40
(x^2*x^3)/x^4
= (x^5)/x^4
= x^(5-4)
= x^1
= x
*********
f(x)=2x-3
To find the inverse of the function, interchange the variables and solve for f^(-1)(x).
x = 2f^(-1)(x)-3
Since f^(-1)(x) is on the right-hand side of the equation, switch the sides so it is on the left-hand side of the equation.
2f^(-1)(x)-3 = x
Move all terms not containing f^(-1)(x) to the right-hand side of the equation.
2f^(-1)(x) = x+3
Divide each term in the equation by 2.
f^(-1)(x) = (x+3)/(2)
********
y+4x = 27
xy+x = 40
Since 4x does not contain the variable to solve for, move it to the right-hand side of the equation by subtracting 4x from both sides.
y = -4x+27
xy+x = 40
Since the equation was solved for y, replace all occurrences of y in the other equations with the solution…
y = -4x+27
x(-4x+27)+x=40
y = -4x+27
(-4x^(2)+27x)+x = 40
y = -4x+27
-4x^(2)+28x=40
To set the left-hand side of the equation equal to 0, move all the expressions to the left-hand side.
y=-4x+27
-4x^(2)+28x-40=0
y=-4x+27
4x^(2)-28x+40=0
Factor out the GCF of 4 from 4x^(2)-28x+40…
y = -4x+27
4(x^(2)-7x+10) = 0
Factor the trinomial…
y = -4x+27
4(x-2)(x-5) = 0
Divide both sides of the equation by 4. Dividing 0 by any non-zero number is 0.
y = -4x+27
(x-2)(x-5) = 0
Set each of the factors of the left-hand side of the equation equal to 0….
y = -4x+27
x-2 = 0
x-5 = 0
y = -4x+2
x = 2
x-5 = 0
Since -5 does not contain the variable to solve for, move it to the right-hand side of the equation by adding 5 to both sides.
y = -4x+27
x = 2
x = 5
The complete solution is the set of the individual solutions.
y = -4x+27
x = 2, 5
Multiply -4 by each term inside the parenthesis.
For x = 2
y = -8+27
For x = 2
y = 19
For x=5_y=-20+27
x = 5
y = 7
The final solution is the set of all solutions.
x = 2, y = 19
x = 5, y = 7
Watch Video on Graphing Quadratic Functions – Algebra Help
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