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3 years ago

6

12(6-2)-1.54(125÷5)-3+6

1 answer:

mojhsa [17]3 years ago

4 0

Hey there! :)

1) 12(6 - 2) - 1.5

Simplify within the parenthesis first.

12(4) - 1.5

Multiply the parenthesis out.

48 - 1.5

Simplify.

46.5

2) 4(125 ÷ 5) - 3 + 6

Simplify within the parenthesis.

4(25) - 3 + 6

Multiply out the parenthesis.

100 - 3 + 6

Simplify.

106 - 3

Simplify.

103

~Hope I helped!~

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Please someone help me...

laiz [17]

Step-by-step explanation:

First factor out the negative sign from the expression and reorder the terms

That's

$\frac{1}{ - (( \tan(2A) - \tan(6A) )} - \frac{1}{ \cot(6A) - \cot(2A) }$

<u>Using trigonometric </u><u>identities</u>

That's

<h3> $\cot(x) = \frac{1}{ \tan(x) }$ </h3>

<u>Rewrite the expression</u>

That's

$\frac{1}{ - (( \tan(2A) - \tan(6A) )} - \frac{1}{ \frac{1}{ \tan(6A) } } - \frac{1}{ \frac{1}{ \tan(2A) } }$

We have

<h3> $- \frac{1}{ \tan(2A) - \tan(6A) } - \frac{1}{ \frac{ \tan(2A) - \tan(6A) }{ \tan(6A) \tan(2A) } }$ </h3>

<u>Rewrite the second fraction</u>

That's

<h3> $- \frac{1}{ \tan(2A) - \tan(6A) } - \frac{ \tan(6A) \tan(2A) }{ \tan(2A) - \tan(6A) }$ </h3>

Since they have the same denominator we can write the fraction as

$- \frac{1 + \tan(6A) \tan(2A) }{ \tan(2A) - \tan(6A) }$

Using the identity

<h3> $\frac{x}{y} = \frac{1}{ \frac{y}{x} }$ </h3>

<u>Rewrite the expression</u>

We have

<h3> $- \frac{1}{ \frac{ \tan(2A) - \tan(6A) }{1 + \tan(6A) \tan(2A) } }$ </h3>

<u>Using the trigonometric identity</u>

<h3> $\frac{ \tan(x) - \tan(y) }{1 + \tan(x) \tan(y) } = \tan(x - y)$ </h3>

<u>Rewrite the expression</u>

That's

<h3> $- \frac{1}{ \tan(2A -6A) }$ </h3>

Which is

<h3> $- \frac{1}{ \tan( - 4A) }$ </h3>

<u>Using the trigonometric identity</u>

<h3> $\frac{1}{ \tan(x) } = \cot(x)$ </h3>

Rewrite the expression

That's

<h3> $- \cot( - 4A)$ </h3>

<u>Simplify the expression using symmetry of trigonometric functions</u>

That's

<h3> $- ( - \cot(4A) )$ </h3>

<u>Remove the parenthesis </u>

We have the final answer as

<h2> $\cot(4A)$ </h2>

As proven

Hope this helps you

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