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3 years ago

For which function is the domain -5 ≤ x ≤ -1 not appropriate? y = √3x y = 3x y = 3x y = x3

1 answer:

4 0

There is no real values for the square root of a negative value, therefore that negative domain does not exist for the function y=√(3x).

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The height, h, in feet of the tip of the hour hand of a wall clock varies from 9 feet to 10 feet. Which of the following equatio

bixtya [17]

The correct option is (b) h=0.5cos( $\pi$ /6t)+9.5.

The equations can be used to model the height as a function of time, t, in hours is h=0.5cos( $\pi$ /6t)+9.5.

<h3>Equation of cosine function:</h3>

The following is a presentation of the cosine function's generic form;

y = a + cos(bx - c) + d

amplitude = a

b = cycle speed

Calculation for the model height;

The height, h (feet) of the tip of the hour hand of a wall clock varies from 9 feet to 10 feet.

Obtain amplitude 'a' as

$\begin{aligned}a &=\frac{\text { Maximum value }-\text { Minimum value }}{2} \\a &=\frac{10-9}{2} \\a &=\frac{1}{2} \\a &=0.5\end{aligned}$

The time 'T' is calculated as-

$\begin{aligned}&\mathrm{T}=\frac{2 \pi}{\mathrm{b}} \\&12=\frac{2 \pi}{\mathrm{b}} \\&\mathrm{b}=\frac{2 \pi}{12} \\&\mathrm{~b}=\frac{\pi}{6}\end{aligned}$

Now, calculate 'd'

$\begin{aligned}&\mathrm{d}=\frac{\text { Maximum value }+\text { Minimum value }}{2} \\&\mathrm{~d}=\frac{10+9}{2} \\&\mathrm{~d}=\frac{19}{2} \\&\mathrm{~d}=9.5\end{aligned}$

Therefore, with the height as a function of time, t, expressed in hours, can be modeled by the following equations:

h=0.5cos( $\pi$ /6t)+9.5

To know more about the general equation of a cosine function, here

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The complete question is-

The height, h, in feet of the tip of the hour hand of a wall clock varies from 9 feet to 10 feet. Which of the following equations can be used to model the height as a function of time, t, in hours? Assume that the time at t = 0 is 12:00 a.m.

A. h=0.5cos( $\pi$ /12t)+9.5

B. h=0.5cos( $\pi$ /6t)+9.5

C. h=cos( $\pi$ /12t)+9

D. h=cos( $\pi$ /6t)+9

7 0

2 years ago

F(x) = x4 - 9x3 - 8x2 - 9x - 2; f(-3)<br> -

NISA [10]

Answer:

Step-by-step explanation:

f(x) = x⁴ - 9x³ - 8x² - 9x - 2

f(-3) = (-3)⁴ -9*(-3)³ - 8*(-3)² - 9*(-3) - 2

= 81 - 9*(-27) - 8*9 + 27 -2

= 81 + 243 - 72 + 27 - 2

= 277

7 0

2 years ago

Read 2 more answers

Which solution to the equation 1/x-1=x-2/2x^2-2 is extraneous?

olga2289 [7]

<span>An extraneous solution is a solution, such as that to an equation, that emerges from the process of solving the problem but is not a valid solution to the original problem.

Given the equation:
$\frac{1}{x-1} = \frac{x-2}{2x^2-2} \\ \\ 2x^2-2=(x-1)(x-2)=x^2-3x+2 \\ \\ x^2+3x-4=0 \\ \\ (x-1)(x+4)=0 \\ \\ x-1=0 \ or \ x+4=0 \\ \\ x=1 \ or \ x=-4$

From the equation, it is obtained that x = 1 and x = -4 are the solutions of the equation.

Now, we substitute x = 1 into the equation as follows:
</span><span> $\frac{1}{1-1} = \frac{1-2}{2(1)^2-2} \\ \\ \frac{1}{0} = \frac{-1}{2-2} \\ \\ \frac{1}{0} = \frac{-1}{0}$

As can be see, for x = 1, the equation is undefined.

Now, we substitute x = -4 into the equation as follows:
$\frac{1}{-4-1} = \frac{-4-2}{2(-4)^2-2} \\ \\ \frac{1}{-5} = \frac{-6}{2(16)-2} = \frac{-6}{32-2} = \frac{-6}{30} \\ \\ - \frac{1}{5} =- \frac{1}{5}$

It can be seen that x = -4 is a valid solution of the orignal equation.

Therefore, x = 1 is an extraneous solution to the equation
$\frac{1}{x-1} = \frac{x-2}{2x^2-2}$ </span>

8 0

3 years ago

True or False

damaskus [11]

Answer:

False.

Step-by-step explanation:

it is an angle formed by 2 adjacent sides inside the polygon.

5 0

4 years ago

Read 2 more answers

Find the perimeter of the polygon. Please help