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2 years ago

How many centimeters are there in 1/5 of an inch

Mathematics

1 answer:

Lynna [10]2 years ago

4 0

Answer:

0.508

Step-by-step explanation:

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Simplify the expression by combining like terms.

Tju [1.3M]

Answer:

8x + 13

Step-by-step explanation:

"Like terms" means the same kind of algebraic expression...the 6x and the 2x are "like terms" because they both have an x in them. The 9 and the 4 are both constants, which means they are plain numbers. So 6x and 2x is 8x .

9+4 is 13 just like elementary school.

So all together,

6x + 9 + 2x + 4

6x + 2x + 9 + 4

8x + 13

We can rearrange the terms because the problem has all addition in between the terms.

4 0

2 years ago

I need to know what the variable of x is please explain

Kisachek [45]

The answer is: x = 36

3 0

3 years ago

Read 2 more answers

What is <br> 2983x + (129x123) -3818<br> x=969

True [87]

Answer:

the answer of it is 2902581

3 0

2 years ago

At Hogwarts School of Witchcraft and Wizardry, Professor Snape was concerned about grade inflation, and suggested that the schoo

ExtremeBDS [4]

Answers:

The z scores are approximately:

Care of Magical Creatures: z = 0.333
Defense Against the Dark Arts: z = 0.583
Transfiguration: z = -0.263
Potions: z = -0.533

From those scores, we can say:

Best grade = Defense Against the Dark Arts
Worst grade = Potions

=====================================================

Further Explanation:

We'll need to convert each given score to a corresponding standardized z score.

The formula to use is

z = (x - mu)/sigma

where,

x = given grade for each class
mu = mean
sigma = standard deviation

Let's find the z score for the Care of Magical Creatures class

z = (x - mu)/sigma

z = (3.80 - 3.75)/(0.15)

z = 0.333 approximately

Repeat this process for the Defense Against the Dark Arts score.

z = (x - mu)/sigma

z = (3.60 - 3.25)/(0.60)

z = 0.583 approximately

And for the Transfiguration class as well

z = (x - mu)/sigma

z = (3.10 - 3.20)/(0.38)

z = -0.263 approximately

The negative z score means his grade below the average, whereas earlier the other scores were above the average since he got positive z scores.

Now do the final class (Potions) to get this z score

z = (x - mu)/sigma

z = (2.50 - 2.90)/(0.75)

z = -0.533 approximately

This grade is below average as well.

----------------------------

To summarize, we have these z scores

Care of Magical Creatures: z = 0.333
Defense Against the Dark Arts: z = 0.583
Transfiguration: z = -0.263
Potions: z = -0.533

Harry did his best in Defense Against the Dark Arts because the z score of 0.583 (approximate) is the largest of the four z scores. On the other hand, his worst grade is in Potions because -0.533 is the lowest z score.

3 0

1 year 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

2 years ago

Read 2 more answers