All categories

3 years ago

Point GG is located at (1,-2)(1,−2) on the coordinate plane. Point GG is reflected over the yy-axis to create point G'G

Mathematics

1 answer:

yan [13]3 years ago

4 0

Step-by-step explanation:

starting at (1,-2) the point will move to the left 2 to be reflected over the y-axis and will then be at (-1,-2). then reflecting over the x-axis will move it up 4 and end up at (-1,2).

You might be interested in

Find the unit rate. 256 miles per 8 gallons

denpristay [2]

<span>the answer is 32 miles per 1 gallon.</span>

3 0

3 years ago

Read 2 more answers

A perfume company claims that the mean weight of ther new perfume is at least 8.9 fluid oz. Express the null hypothesis and the

zlopas [31]

Answer:

$H0: mu = 8.9 fl oz.\\\\$

$Ha: mu ≠8.9 fluid oz$

Step-by-step explanation:

Given that A perfume company claims that the mean weight of ther new perfume is at least 8.9 fluid oz

For testing this claim, in Statistics we perform a certain measures called hypothesis testing.

For this first step is to create null and alternate hypothesis.

Normally null hypothesis would have some statistic = something

Here we want to test the mean weight of perfume

Hence null hypothesis would be

H0: mu = 8.9 fl oz.

Alternate hypothesis would be opposite of this claim

i.e.

Ha: mu ≠8.9 fluid oz

Hence answer is

$H0: mu = 8.9 fl oz.\\\\$

$Ha: mu ≠8.9 fluid oz$

6 0

3 years ago

What is the interquartile range of the following data set?3,8,14,19,22,29,33,37,43,49

Vaselesa [24]

Q1: 14
Q3: 37
Interquartile: 37-14 = 23

THE ANSWER IS 23

(Brainliest please?)

4 0

3 years ago

Find the limit

Lana71 [14]

Step-by-step explanation:

<h3>Appropriate Question :-</h3>

Find the limit

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

On substituting directly x = 1, we get,

$\rm \: = \: \sf \dfrac{1-2}{1 - 1}-\dfrac{1}{1 - 3 + 2}$

$\rm \: = \sf \: \: - \infty \: - \: \infty$

which is indeterminant form.

Consider again,

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right]$

can be rewritten as

$\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 3x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( {x}^{2} - 2x - x + 2)}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x( x(x - 2) - 1(x - 2))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x(x - 1)}-\dfrac{1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ {(x - 2)}^{2} - 1}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 2 - 1)(x - 2 + 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)(x - 1)}{x(x - 2) \: (x - 1))}\right]$

$\rm \: = \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{ (x - 3)}{x(x - 2)}\right]$

$\rm \: = \: \sf \: \dfrac{1 - 3}{1 \times (1 - 2)}$

$\rm \: = \: \sf \: \dfrac{ - 2}{ - 1}$

$\rm \: = \: \sf \boxed{2}$

Hence,

$\rm\implies \:\boxed{ \rm{ \:\rm \: \sf {\displaystyle{\lim_{x\to 1}}} \: \left[\dfrac{x-2}{x^2-x}-\dfrac{1}{x^3-3x^2+2x}\right] = 2 \: }}$

$\rule{190pt}{2pt}$

7 0

3 years ago

Read 2 more answers

Write the expression with fewer terms. <br><br> 10−2

fiasKO [112]

10-2 = 8
2^3 =8
(2x2x2) = 8

7 0

3 years ago

Read 2 more answers