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liq [111]
2 years ago
14

The length of a rectangle is 5 in. More than 3 times its width. The quadratic function f(x) = x(3x + 5) represents the area of t

he rectangle in terms of its width. What is a reasonable domain of the function?
Mathematics
1 answer:
True [87]2 years ago
6 0

Answer: Domain of function is R^{ + } or (0, \infty )

Step-by-step explanation:

Assuming x as width of rectangle

so, Width = x

Question says, Length is 5in more than 3 times id width.

we write as Length = 3x+5

Therefore, Area of rectangle will be given function f(x)

f(x) = x(3x+5)

Here, It is important to notice that width and length of rectangle will always be positive values and also, area of rectangle is always positive.

we can write in equation as

x > 0

f(x) > 0

Thus, Domain of function is R^{ + } or (0, \infty )

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In this diagram, BAC~EDF. If the area of BAC=15 in2, what is the area of EDF? 4, 5.
lesya692 [45]

Answer:

9.6 square inches.

Step-by-step explanation:

We are given that ΔBAC is similar to ΔEDF, and that the area of ΔBAC is 15 inches. And we want to determine the area of ΔDEF.

First, find the scale factor <em>k</em> from ΔBAC to ΔDEF:

\displaystyle 5 k = 4

Solve for the scale factor <em>k: </em>

<em />\displaystyle k = \frac{4}{5}<em />

<em />

Recall that to scale areas, we square the scale factor.

In other words, since the scale factor for sides from ΔBAC to ΔDEF is 4/5, the scale factor for its area will be (4/5)² or 16/25.

Hence, the area of ΔEDF is:

\displaystyle \Delta EDF = \frac{16}{25}\left(15\right) = 9.6\text{ in}^2

In conclusion, the area of ΔEDF is 9.6 square inches.

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