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2 years ago

Find the length of AC on this triangle

Mathematics

1 answer:

Illusion [34]2 years ago

4 0

Answer:

Step-by-step explanation:

Using the tangent ratio in the right triangle

tan12° = $\frac{opposite}{adjacent}$ = $\frac{AC}{BC}$ = $\frac{AC}{44}$ ( multiply both sides by 44 )

44 × tan12° = AC , then

AC ≈ 9.35 ( to 2 dec. places )

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2 1/5 × 5/6 (its a fraction) PLSSS i need it now its a test

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Answer:

$\frac{11}{6} \ \ or \ 1\frac{5}{6}$

Step-by-step explanation:

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If a/b < c/d with b > 0, d > 0, prove that a+c / b+d lies between the two fractions a/b and c/d .

Tems11 [23]

Divide through everything by b :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ab + \dfrac cb}{1 + \dfrac db}$

Since a/b < c/d, it follows that

$\dfrac{a+c}{b+d} < \dfrac{\dfrac cd+\dfrac cb}{1 + \dfrac db}$

Multiply through everything on the right side by b/d to get

$\dfrac{a+c}{b+d} < \dfrac{\dfrac{bc}{d^2}+\dfrac cd}{\dfrac bd+1} = \dfrac{\dfrac cd\left(\dfrac bd+1\right)}{\dfrac bd+1} = \dfrac cd$

and so (a + c)/(b + d) < c/d.

For the other side, you can do something similar and divide through everything by d :

$\dfrac{a+c}{b+d} = \dfrac{\dfrac ad + \dfrac cd}{\dfrac bd + 1}$

and a/b < c/d tells us that

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ad + \dfrac ab}{\dfrac bd + 1}$

Then

$\dfrac{a+c}{b+d} > \dfrac{\dfrac ab + \dfrac{ad}{b^2}}{1 + \dfrac db} = \dfrac{\dfrac ab\left(1+\dfrac db\right)}{1 + \dfrac db} = \dfrac ab$

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2 years ago

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1/6^8

Step-by-step explanation:

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1 year ago

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Fed [463]

Answer:

The answer is A

Step-by-step explanation:

Supplementary angles are angles which add up to 180°

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