What numbers are a distance of 4\6 unit from 3\6 on a number line?Please help me out.

Mathematics

1 answer:

sweet [91]2 years ago

3 0

The numbers that are a distance of 4/6 unit from 3/6 on a number line are -1/6 and 7/6

<h3>What is the distance on a Number Line?</h3>

We want to know what numbers are a distance of 4/6 unit from 3/6 on a number line.

Add 4/6 to 3/6, and also subtract 4/6 from 3/6. That will give the two numbers.

3/6 + 4/6 = 3/6 + 4/6 = 7/6

3/6 - 4/6 = -1/6

Thus, the numbers that are a distance of 4/6 unit from 3/6 on a number line are -1/6 and 7/6

Read more about Distance on a number Line at; brainly.com/question/17143316

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<h2>Greetings!</h2><h3>Firstly, the slope is found by rearranging the formula:</h3>

3y = 4x + 39

y = $\frac{4x}{3}$ + $\frac{39}{3}$

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<h3>Next, the slope is simply the number in front of the x, </h3>

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tatyana61 [14]

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$\cos { \left( F \right) } =\frac { { e }^{ 2 }+{ g }^{ 2 }-{ f }^{ 2 } }{ 2eg }$

Therefore:

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But what is e?

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And:

$\frac { e }{ \sin { \left( E \right) } } =\frac { g }{ \sin { \left( G \right) } }$

Which means that:

$\frac { e }{ \sin { \left( 76 \right) } } =\frac { 6 }{ \sin { \left( 32 \right) } } \\ \\ \therefore \quad e=\frac { 6\cdot \sin { \left( 76 \right) } }{ \sin { \left( 32 \right) } }$

If you take this value into account, you will discover that f is...

$f=\sqrt { \frac { 6\cdot \sin { \left( 76 \right) } }{ \sin { \left( 32 \right) } } \left( \frac { 6\cdot \sin { \left( 76 \right) } }{ \sin { \left( 32 \right) } } -12\cos { \left( 72 \right) } \right) +36 } \\ \\ \therefore \quad f=10.8\quad \left( 1\quad d.p \right)$

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