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2 years ago

Find the domain of the following piecewise function

Mathematics

1 answer:

r-ruslan [8.4K]2 years ago

6 0

The domain of the function will be [-4, 6). Then the correct option is B.

The complete question is attached below.

<h3>What are domain and range?</h3>

The domain means all the possible values of x and the range means all the possible values of y.

The function is given below.

$f(x) = \left\{\begin{matrix}x+4, & if & -4\leq x < 3 \\\\2x-1, & if & 3 \leq x < 6 \\\end{matrix}\right.$

Then the domain of the function will be [-4, 6).

Then the correct option is B.

More about the domain and range link is given below.

brainly.com/question/12208715

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A city had population 67,255 on january 1, 2000, and its population has been increasing by 2935 people each year since then. A l

irina1246 [14]

We are given: On january 1, 2000 initial population = 67,255.

Number of people increase each year = 2935 people.

Therefore, 67,255 would be fix value and 2935 is the rate at which population increase.

Let us assume there would be t number of years after year 2000 and population P after t years is taken by function P(t).

So, we can setup an equation as

Total population after t years = Number of t years * rate of increase of population + fix given population.

In terms of function it can be written as

P(t) = t * 2935 + 67255.

Therefore, final function would be

P(t) = 2935t +67255.

So, the correct option is d.P(t) = 67255 + 2935t.

4 0

3 years ago

Read 2 more answers

If we inscribe a circle such that it is touching all six corners of a regular hexagon of side 10 inches, what is the area of the

Brrunno [24]

Answer:

$\left(100\pi - 150\sqrt{3}\right)$ square inches.

Step-by-step explanation:

<h3>Area of the Inscribed Hexagon</h3>

Refer to the first diagram attached. This inscribed regular hexagon can be split into six equilateral triangles. The length of each side of these triangle will be $10$ inches (same as the length of each side of the regular hexagon.)

Refer to the second attachment for one of these equilateral triangles.

Let segment $\sf CH$ be a height on side $\sf AB$ . Since this triangle is equilateral, the size of each internal angle will be $\sf 60^\circ$ . The length of segment

$\displaystyle 10\, \sin\left(60^\circ\right) = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3}$ .

The area (in square inches) of this equilateral triangle will be:

$\begin{aligned}&\frac{1}{2} \times \text{Base} \times\text{Height} \\ &= \frac{1}{2} \times 10 \times 5\sqrt{3}= 25\sqrt{3} \end{aligned}$ .

Note that the inscribed hexagon in this question is made up of six equilateral triangles like this one. Therefore, the area (in square inches) of this hexagon will be:

$\displaystyle 6 \times 25\sqrt{3} = 150\sqrt{3}$ .

<h3>Area of of the circle that is not covered</h3>

Refer to the first diagram. The length of each side of these equilateral triangles is the same as the radius of the circle. Since the length of one such side is $10$ inches, the radius of this circle will also be $10$ inches.

The area (in square inches) of a circle of radius $10$ inches is:

$\pi \times (\text{radius})^2 = \pi \times 10^2 = 100\pi$ .