Answer: f(x^-1) = x/5 - 3/5
Step-by-step explanation:
1. Replace f(x) with y
2. Swap the positions of x and y to make x = 5y + 3
3. Solve for y by subtracting 3 from both sides and dividing each side by 5
Answer:
Total amount of fencing needed as an algebraic expression in terms of x is: <em>10x</em><em> </em><em>+</em><em> </em><em>3</em> .
Step-by-step explanation:
As it is given that each rectangle has the same dimensions, the dimensions of each rectangle must be: x units by 2x + 1 units.
Based on this, we can calculate the total amount of fencing needed.
Let width of each rectangle = x
Let length of each rectangle = 2x + 1
There are 4 widths and 3 lengths in total of fencing.
Therefore:
= 4 ( x ) + 3 ( 2x + 1 )
Expand:
= 4x + 6x + 3
Group like-terms:
= 10x + 3
The answer is 2X^3+12x^2+10x-24
For problems like this, change the wording into simple algebraic
formulas. We know that the area of a rectangle is its width times its
length, or A=LW. The problem tells you that the area is 48, so LW=48.
Also, if the length = twice the width minus 4, then L=2W-4.
From here, substitute your new L value for the L in the first equation:
48=LW --> 48=(2W-4)W
Now solve this equation by factoring it out into a polynomial:
48=2W^2-4W --> 24=W^2-2W --> W^2-2W-24=0 --> (W+4)(W-6)=0.
Solving this equation gives us W values of 6 and -4, but since the width
of a rectangle cannot be negative, the width must be 6.
Since L=2W-4, then L=2(6)-4 --> L=12-4 = 8. Therefore, the dimensions are 6X8, or 6 feet by 8 feet.
To check our work: The length equals four feet less than twice the width: 8=2(6)-4 --> 8=12-4 --> 8=8. This checks out.
Also, the area is 48 ft^2, and (6)(8) = 48, so this also checks out.
Hopefully this helps. If you need any more help or if I went over something too quickly, just let me know.
