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3 years ago

6

Can you tell me the answers and how you got them??thx

1 answer:

MAXImum [283]3 years ago

8 0

Use the distance and midpoint formula. Distance formula is: square root of (x2-x1)^2 + (y2-y1)^2. Midpoint formula is: (x1 +x2)/2 , (y2+y1)/2.

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Quota sampling is most commonly used in a. descriptive research. b. collecting primary data. c. surveys. d. population research.

3241004551 [841]

Answer:

Quota sampling method is a non-probability sampling and it can be defined as a sampling method of gathering representative data from a group. Application of quota sampling ensures that sample group represents certain characteristics of the population chosen by the researcher

Step-by-step explanation:

7 0

3 years ago

Describe the explicit formula?

lukranit [14]

Answer:

its to a power so it is an exponential power

5 0

3 years ago

3xy=9 is it direct variation or inverse variation or neither

steposvetlana [31]

$3xy = 9$

$y = \dfrac 9 {3x}$

$y = \dfrac 3 x$

y varies inversely with x

Answer: inverse variation

5 0

3 years ago

Please help me slove this problem <br> 3x - 3 = 12

allsm [11]

The answer is 5! Hope this helps!

8 0

3 years ago

Read 2 more answers

Determine the domain of the function (fog) (x) where f(x) = 3x-1/x-4 and g(x)=x+1/x

Mrac [35]

Answer:

$(-\infty,0)\cup(0, \frac{1}{3})\cup (\frac{1}{3},\infty)$

Step-by-step explanation:

The given functions are $f(x)=\frac{3x-1}{x-4}$ and $g(x)=\frac{x+1}{x}$ .

We now composed the two functions to find:

$(f\circ g)(x)=f(g(x))$

$\implies (f\circ g)(x)=f(\frac{x+1}{x})$

$\implies (f\circ g)(x)=\frac{3(\frac{x+1}{x})+1}{\frac{x+1}{x}-4}$

$\implies (f\circ g)(x)=\frac{4x+3}{1-3x}$

This function is defined if the denominator is not zero.

$1-3x\ne0$

$x\ne\frac{1}{3}$

We write this in interval notation as:

$(-\infty,\frac{1}{3})\cup (\frac{1}{3},\infty)$

We need to be cautious here as x=0 is not in the domain of g(x).

Therefore the domain of $(f\circ g)(x)$ is

$(-\infty,0)\cup(0, \frac{1}{3})\cup (\frac{1}{3},\infty)$

4 0

3 years ago