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3 years ago

Common denominator and strategy 2/3 and 10/15

2 answers:

8 0

The common denominator of 2/3 and 10/15 is 15.
To go further,
2/3 can be multipled by 5.
So 2 • 5 = 10
And 5 • 3 = 15
Therefore the new fraction is 10/15
When comparing to the 10/15, both of them are equivalent fractions.

Tom [10]3 years ago

3 0

Their common denominator is 15. Hope it helps! :) If you could vote my answer as the brainliest, that would be awesome! :)

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Answer:

$\sqrt{10} \: \: units$

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NEED ASAP plz. I will also mark brainiest! Don't just take points, I will report you.

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Part A:

You may choose the two lines connecting the origin and points A and B, and choose the portion of the space between them.

The line between the origin and A is

$y = 3x$

We want everything below this line (line included), so the first inequality is

$y \leq 3x$

The line between the origin and B is

$y = \dfrac{1}{3}x$

We want everything above this line (line included), so the second inequality is

$y \geq \dfrac{1}{3}x$

Create a system with these two inequalities and you'll have an area including only points A and B

Part B:

To verify the solutions, we can plug the coordinates of A and B in this system and check that we get something true: the coordinates of point A are (1,3), while the coordinates of point B are (3,1). The system becomes:

$A:\begin{cases}3 \leq 3\cdot 1\\3 \geq \frac{1}{3}\cdot 1\end{cases},\quad B:\begin{cases}1 \leq 3\cdot 3\\1 \geq \frac{1}{3}\cdot 3\end{cases}$

Which means

$A:\begin{cases}3 \leq 3\\3 \geq \frac{1}{3}\end{cases},\quad B:\begin{cases}1 \leq 9\\1 \geq 1\end{cases}$

And these are all true. So, the system is satisfied, which means that the points belong to the shaded area.

Part C

If you draw the line, you'll see that the only points that lay below the line are B and C. In fact, if we plug the coordinates we have

$B:\ 1$

And this are both true. You can check the coordinates of all other points, and see that they won't satisfy the inequality y<3x-6

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Answer:

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