Perimeter of rectangle is 8x and Area is 3x^2 + 2x -1..
Lets try to solve out the problem,
Length = 3x-1
Width = x+1
Perimeter = 2(l+b)
Area = l*b
Perimeter= 2 (3x-1 + x+1)
P= 8x
Area = (3x-1) (x+1)
=> 3x(x+1) -1 (x+1)
=> 3x^2 + 3x -x -1
=> 3x^2 + 2x -1.
Hence Perimeter of rectangle is 8x and Area is 3x^2 + 2x -1..
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Answer:
Step-by-step explanation:
-3 + 1/2n = 1/2 (-n + 14)
distribute
-3 + 1/2n = -1/2n + 7
-1/2n
cancel out -1/2n
-3 = n + 7
-10 = n
subtract 7
n = -10
The points at which a quadratic equation intersects the x-axis are referred to as x intercepts or zeros or roots of quadratic equation
Given :
The points at which a quadratic equation intersects the x-axis
The points at which the any quadratic equation crosses or touches the x axis are called as x intercepts.
At x intercepts the value of y is 0.
So , the points at which a quadratic equation intersects the x-axis is also called as zeros or roots of the quadratic equation .
The points at which a quadratic equation intersects the x-axis are referred to as x intercepts or zeros or roots of quadratic equation
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