Asked by kelly
                Solve the system of equations
3x-y=11
-5x+y=12
2.
Perform the indicated operation and simplify (5/(x-2)+(4(x^2-2x)
3.simpify and write with only positive exponents (7x^3y^-2)/(x^-4y^9)^-3
4.
find the solution of the inequality
2x-11 is less than or equal to -4(5x-3)
5. the length of a recangle is 5 feet less than twice the width. the area is 25 square feet. using w as the variable. write an equation that can be used to calculate the width.
(2w^2-5)
            
        3x-y=11
-5x+y=12
2.
Perform the indicated operation and simplify (5/(x-2)+(4(x^2-2x)
3.simpify and write with only positive exponents (7x^3y^-2)/(x^-4y^9)^-3
4.
find the solution of the inequality
2x-11 is less than or equal to -4(5x-3)
5. the length of a recangle is 5 feet less than twice the width. the area is 25 square feet. using w as the variable. write an equation that can be used to calculate the width.
(2w^2-5)
Answers
                    Answered by
            kelly
            
    Any help would be appreciated. This is for my study guide in order to do well on my test. Someone please help.
    
                    Answered by
            Steve
            
    #1
From the first, y=3x-11
Plug that into the second:
-5x+3x-11 = 12
-2x = 23
x = -23/2
y = -91/2
Or, you can just add the equations to eliminate the y terms:
-5x = 23
...
#2 Your parentheses are unbalanced. If you mean
5/(x-2) + 4/(x^2-2x)
5/(x-2) + 4/(x(x-2))
5x/(x(x-2)) + 4/(x(x-2))
(5x+4)/(x(x-2))
(5x+4)/(x^2-2x)
#3
(7x^3y^-2)/(x^-4y^9)^-3
(x^-4y^9)^-3 = x^12 y^-27
Since you're dividing by that, you have
(7x^3y^-2)*(x^-12 y^27)
7x^-9 y^25
7y^25/x^9
    
From the first, y=3x-11
Plug that into the second:
-5x+3x-11 = 12
-2x = 23
x = -23/2
y = -91/2
Or, you can just add the equations to eliminate the y terms:
-5x = 23
...
#2 Your parentheses are unbalanced. If you mean
5/(x-2) + 4/(x^2-2x)
5/(x-2) + 4/(x(x-2))
5x/(x(x-2)) + 4/(x(x-2))
(5x+4)/(x(x-2))
(5x+4)/(x^2-2x)
#3
(7x^3y^-2)/(x^-4y^9)^-3
(x^-4y^9)^-3 = x^12 y^-27
Since you're dividing by that, you have
(7x^3y^-2)*(x^-12 y^27)
7x^-9 y^25
7y^25/x^9
                    Answered by
            Steve
            
    #4.
2x-11 <= -4(5x-3)
2x-11 <= -20x + 12
22x <= 23
x <= 23/22
#5.
length is 2w-5
area is 25, so
w(2w-5) = 25
2w^2-5w = 25
2w^2 - 5w - 25 = 0
(2w+5)(w-5) = 0
w=5 (-5/2 is not a usable width)
the rectangle is a square 5x5
    
2x-11 <= -4(5x-3)
2x-11 <= -20x + 12
22x <= 23
x <= 23/22
#5.
length is 2w-5
area is 25, so
w(2w-5) = 25
2w^2-5w = 25
2w^2 - 5w - 25 = 0
(2w+5)(w-5) = 0
w=5 (-5/2 is not a usable width)
the rectangle is a square 5x5
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