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seraphim [82]
3 years ago
14

What is the absolute value of -7.5?

Mathematics
2 answers:
Artemon [7]3 years ago
8 0

Answer:

3

Step-by-step explanation:

Iteru [2.4K]3 years ago
8 0

Answer:

I don't really understand the question but the absolute values of everything is

7.5

10

9

6

4

2

0

2

5

6

8

10

7.5

Step-by-step explanation:

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Could I get help on this question in fraction form
viktelen [127]

Answer:

\frac{72-\sqrt{323} }{90}

Step-by-step explanation:

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3 years ago
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70% of 120 = 120x70/100 <br> How do you figure this out
Sati [7]
10% of 120 = 12
12 x 7 = 84
70% = 84

120 x 70 = 8400
8400 / 100 = 84

It’s either showing you that both sides of the equation equal to the same thing or that you can find 70% of 120 by using the steps on the right
3 0
3 years ago
Can someone help me pls
Drupady [299]

Answer:

150 degrees

Step-by-step explanation:

75 -----> 75 out of 360

360-75=285

285-75 will be divided by 2

210 divided by 2 equals 105

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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
2 years ago
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I will give u brainlyist!! pls help
Karolina [17]

Answer:

i believe C

Step-by-step explanation:

5 0
3 years ago
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