Does anyone know the answer to this question ?

Is the relation a function?

son4ous [18]

Answer:

No

Steps- two rang values exist for domain value -2

8 0

2 years ago

Use the function values for f and g shown in the table below to evaluate the following expression.

Vikentia [17]

The function f(g(2)) is an illustration of a composite function

The value of f(g(2)) is 2

<h3>How to determine the value of the function f(g(2))?</h3>

Given:

The table of values for functions f(x) and g(x)

To calculate f(g(2)), we start by calculating g(2)

From the table;

g(2) = 6

So, we have:

f(g(2)) = f(6)

From the table;

f(6) = 2

So, we have:

f(g(2)) = 2

Hence, the value of f(g(2)) is 2

Read more about composite functions at:

brainly.com/question/10687170

5 0

2 years ago

A curve in polar coordinates is given by: r=9+3cosθ. Point P is at θ=21π/18 .?

ikadub [295]

Answer:

Step-by-step explanation:

Given that a curve in polar coordinates is given by:

r=9+3cosθ

a) At point P, we have

$θ=\frac{21\pi}{18}$

Substitute to get

$r=9+cos \frac{21\pi}{18}\\=9.3932$

b) Cartesian coordinate is

$x= rcos \theta=3.6934\\y =r sin \tjeta =8.6365$

c) At the origin r =0

when r =0

we have

$9+3cos\theta=0\\cos\theta =-3$

Since cos cannot take values as -3 it doe snot pass through origin.

6 0

2 years ago

a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-

Tresset [83]

Answer:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

$M = \frac{1}{2}(Lower + Upper)$

Where

$Lower \to$ Lower class interval

$Upper \to$ Upper class interval

So, we have:

Class 63-65:

$M = \frac{1}{2}(63 + 65) = 64$

Class 66 - 68:

$M = \frac{1}{2}(66 + 68) = 67$

When the computation is completed, the frequency distribution will be:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

$\bar x = 70.2903$