Assign the following variables for the origina3l rectangle:
let w = width let w + 8 = length and the area would be w(w + 8) = w² + 8w
No for the second rectangle:
let (w + 4) = width and (w + 8 - 5) or (w + 3) = length
Area = length x width or (w + 4)(w + 3) = w² + 3w + 4w + 12 using the foil method to multiply to binomials. Simplified Area = w² + 7w + 12
Now our problem says that the two area will be equal to each other, which sets up the following equation:
w² + 8w = w² + 7w + 12 subtract w² from both sides
8w = 7w + 12 subtract 7w from both sides
w = 12 this is the width of our original rectangle
recall w + 8 = length, so length of the original rectangle would be 20
Answer:
We can't solve this because we can't add graphs. One way you could solve it would using Desmos.
Step-by-step explanation:
3000 is the freaking right answer
9514 1404 393
Answer:
-3 ≤ x ≤ 19/3
Step-by-step explanation:
This inequality can be resolved to a compound inequality:
-7 ≤ (3x -5)/2 ≤ 7
Multiply all parts by 2.
-14 ≤ 3x -5 ≤ 14
Add 5 to all parts.
-9 ≤ 3x ≤ 19
Divide all parts by 3.
-3 ≤ x ≤ 19/3
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<em>Additional comment</em>
If you subtract 7 from both sides of the given inequality, it becomes ...
|(3x -5)/2| -7 ≤ 0
Then you're looking for the values of x that bound the region where the graph is below the x-axis. Those are shown in the attachment. For graphing purposes, I find this comparison to zero works well.
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For an algebraic solution, I like the compound inequality method shown above. That only works well when the inequality is of the form ...
|f(x)| < (some number) . . . . or ≤
If the inequality symbol points away from the absolute value expression, or if the (some number) expression involves the variable, then it is probably better to write the inequality in two parts with appropriate domain specifications:
|f(x)| > g(x) ⇒ f(x) > g(x) for f(x) > 0; or -f(x) > g(x) for f(x) < 0
Any solutions to these inequalities must respect their domains.
The answer of the question is 75