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3 years ago

I need help with this

Mathematics

1 answer:

Svetradugi [14.3K]3 years ago

3 0

Answer:

A, E, and F

Step-by-step explanation:

"Irrational Number. An irrational number is a number that cannot be expressed as a fraction for any integers and. . Irrational numbers have decimal expansions that neither terminate nor become periodic." quoted from mathworld.wolfram.com.

Hope this helps! Have a great day! :)

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Special Triangles

Ugo [173]

Answer:

see explanation

Step-by-step explanation:

Using the cosine and tangent trigonometric ratios and the exact values

cos30° = $\frac{\sqrt{3} }{2}$ and tan30° = $\frac{1}{\sqrt{3} }$ , then

cos30° = $\frac{adjacent}{hypotenuse}$ = $\frac{6}{m}$ = $\frac{\sqrt{3} }{2}$ ( cross- multiply )

$\sqrt{3}$ × m = 12 ( divide both sides by $\sqrt{3}$ )

m = $\frac{12}{\sqrt{3} }$ ← rationalise by multiplying numerator/ denominator by $\sqrt{3}$ )

m = $\frac{12}{\sqrt{3} }$ × $\frac{\sqrt{3} }{\sqrt{3} }$ = $\frac{12\sqrt{3} }{3}$ = 4 $\sqrt{3}$

-------------------------------------------------------------------------------

tan30° = $\frac{opposite}{adjacent}$ = $\frac{n}{6}$ = $\frac{1}{\sqrt{3} }$ ( cross- multiply )

$\sqrt{3}$ × n = 6 ( divide both sides by $\sqrt{3}$ )

n = $\frac{6}{\sqrt{3} }$ ← rationalise the denominator

n = $\frac{6}{\sqrt{3} }$ × $\frac{\sqrt{3} }{\sqrt{3} }$ = $\frac{6\sqrt{3} }{3}$ = 2 $\sqrt{3}$

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