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:
Step-by-step explanation:
Remembert that, by definition:
→
Then, you can rewrite in exponential form:
Now you can solve for the variable "x":
Add 6 to both sides of the equation:
And finally you must divide both sides of the equation by 2, then:
It's clear that for x not equal to 4 this function is continuous. So the only question is what happens at 4.
<span>A function, f, is continuous at x = 4 if
</span><span>
</span><span>In notation we write respectively
</span>
Now the second of these is easy, because for x > 4, f(x) = cx + 20. Hence limit as x --> 4+ (i.e., from above, from the right) of f(x) is just <span>4c + 20.
</span>
On the other hand, for x < 4, f(x) = x^2 - c^2. Hence
Thus these two limits, the one from above and below are equal if and only if
4c + 20 = 16 - c²<span>
Or in other words, the limit as x --> 4 of f(x) exists if and only if
4c + 20 = 16 - c</span>²
That is to say, if c = -2, f(x) is continuous at x = 4.
Because f is continuous for all over values of x, it now follows that f is continuous for all real nubmers
Answer:
4
Step-by-step explanation:
What you want to do in dilation is multiply the side(s) by the scal factor to get the measurements of the new shape. in this case multiply 8 by 3.6 to get the new side length of 28.8