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Igoryamba
3 years ago
15

Harmonic mean if a number h such that (h-a)/(b-h)=a/b Prove h is H(a,b) iff satisfies either relation a. (1/a)-(1/h) = (1/h) - (

1/b) b. h = (2ab)/(a+b)
Mathematics
1 answer:
Fudgin [204]3 years ago
7 0
A.

\displaystyle\frac1a-\frac1h=\frac1h-\frac1b
\implies\displaystyle\frac{h-a}{ah}=\frac{b-h}{bh}
\implies\displaystyle\frac{h-a}{b-h}=\frac{ah}{bh}
\implies\displaystyle\frac{h-a}{b-h}=\frac ab

b.

\displaystyle h=\frac{2ab}{a+b}
\displaystyle\implies\frac{h-a}{b-h}=\frac{\frac{2ab}{a+b}-a}{b-\frac{2ab}{a+b}}
\displaystyle\implies\frac{h-a}{b-h}=\frac{2ab-a(a+b)}{b(a+b)-2ab}
\displaystyle\implies\frac{h-a}{b-h}=\frac{ab-a^2}{b^2-ab}
\displaystyle\implies\frac{h-a}{b-h}=\frac{a(b-a)}{b(b-a)}
\displaystyle\implies\frac{h-a}{b-h}=\frac ab

The other direction can be proved by following the manipulations in the reverse order.
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See explanation

Step-by-step explanation:

16. Two parallel lines are cut by transversal. Angles with measures (6x+20)^{\circ} and (x+100)^{\circ} are alternate exterior angles. By alternate exterior angles, the measures of alternate exterior angles are the same:

6x+20=x+100\\ \\6x-x=100-20\\ \\5x=80\\ \\x=16

Then

(6x+20)^{\circ}=(6\cdot 16+20)^{\circ}=116^{\circ}\\ \\(x+100)^{\circ}=(16+100)^{\circ}=116^{\circ}

17. Two parallel lines are cut by transversal. Angles with measures (2x+12)^{\circ} and (3x-22)^{\circ} are alternate interior angles. By alternate interior angles, the measures of alternate interior angles are the same:

2x+12=3x-22\\ \\2x-3x=-22-12\\ \\-x=-34\\ \\x=34

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18. Two parallel lines are cut by transversal. Angles with measures (6x-7)^{\circ} and (5x+10)^{\circ} are alternate exterior angles. By alternate interior angles, the measures of alternate exterior angles are the same:

6x-7=5x+10\\ \\6x-5x=10+7\\ \\x=17

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19. The diagram shows two complementary angles with measures 2x^{\circ} and 56^{\circ}. The measures of complementary angles add up to 90^{\circ}, then

2x+56=90\\ \\2x=90-56\\ \\2x=34\\ \\x=17

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Check:

34^{\circ}+56^{\circ}=90^{\circ}

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m\angle 1=(5x+7)^{\circ}=(5\cdot 4+7)^{\circ}=27^{\circ}\\ \\m\angle 2=(3x+15)^{\circ}=(3\cdot 4+15)^{\circ}=27^{\circ}

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