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Anna71 [15]
3 years ago
7

Express 0.1345345... using bar notation​

Mathematics
2 answers:
Drupady [299]3 years ago
8 0

Answer:

3/1345

Step-by-step explanation:

Liono4ka [1.6K]3 years ago
7 0

Answer:

0.1345

Step-by-step explanation:

The bolded numbers are what is barred

345 is constantly repeated so that is what would be barred as that repeats for ever

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The ardered.pairs.(0, 1), (3, 10) and (4, n) are solutions of the same linear equation. Find n.
Fittoniya [83]

Answer:

n = 13.

Step-by-step explanation:

Slope of the line = (10-1)/3-0) = 3

So the equation of the line is:

y - 1 = 3(x - 0)

y = 3x + 1

When x = 4 y = n,  so:

n = 3(4) + 1 = 13.

n = 13.

7 0
3 years ago
You stopped on your way to basketball practice and bought four cans of fruit punch for $0.69 each. How much did you spend in all
blondinia [14]

Answer:

0.69 x 4= $2.76

Step-by-step explanation:

4 0
3 years ago
Please solve quickly!!
tamaranim1 [39]

Answer:ya

Step-by-step explanation:

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6 0
3 years ago
A plane headed due south is traveling with an airspeed of 160 miles per
lord [1]

Answer:

60 °

Step-by-step explanation:

Applying the law of sine we have the following:

160 / Sin B = 80 / Sin 30 °

We solve for B and we have:

160 / Sin B = 80 / 0.5

160 / Sin B = 160

Without B = 160/160

B = Arc Sin (1)

B = 90 °

Now to calculate the other angle (true

 course) would be:

A = 180 ° - (B + C)

A = 180 ° - (90 ° + 30 °)

A = 60 °

That is to say that true course is 60 °

5 0
3 years ago
Differentiate<br>y=tan (x^2-5x+6)​
slavikrds [6]

Use the chain rule:

<em>y</em> = tan(<em>x</em> ² - 5<em>x</em> + 6)

<em>y'</em> = sec²(<em>x</em> ² - 5<em>x</em> + 6) × (<em>x</em> ² - 5<em>x</em> + 6)'

<em>y'</em> = (2<em>x</em> - 5) sec²(<em>x</em> ² - 5<em>x</em> + 6)

Perhaps more explicitly: let <em>u(x)</em> = <em>x</em> ² - 5<em>x</em> + 6, so that

<em>y(x)</em> = tan(<em>x</em> ² - 5<em>x</em> + 6)   →   <em>y(u(x))</em> = tan(<em>u(x)</em> )

By the chain rule,

<em>y'(x)</em> = <em>y'(u(x))</em> × <em>u'(x)</em>

and we have

<em>y(u)</em> = tan(<em>u</em>)   →   <em>y'(u)</em> = sec²(<em>u</em>)

<em>u(x)</em> = <em>x</em> ² - 5<em>x</em> + 6   →   <em>u'(x)</em> = 2<em>x</em> - 5

Then

<em>y'(x)</em> = (2<em>x</em> - 5) sec²(<em>u</em>)

or

<em>y'(x)</em> = (2<em>x</em> - 5) sec²(<em>x</em> ² - 5<em>x</em> + 6)

as we found earlier.

5 0
3 years ago
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