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valentina_108 [34]

3 years ago

9

1. What is the minimum number of vertices in a binary tree with height 5? What is the maximum number of vertices in a binary tre

e with height 5? See Chapter 4.4.

1 answer:

Hunter-Best [27]3 years ago

6 0

Answer:

Height = 5

Minimum number of vertices in a binary tree

whose height is h.

So, there must be At least one node at each of first h levels.

Minimum number of vertices = h

So, Minimum number of vertices of height 5 is 5

The maximum number of nodes in a binary tree of height h = $2^{h+1}- 1$

Substitute h = 5

The maximum number of nodes in a binary tree of height 5 = $2^{5+1}- 1$

The maximum number of nodes in a binary tree of height 5 = 63

So, the maximum number of vertices in a binary tree with height 5 is 63

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Find all the missing dimensions when a=9, b=13, c=64°

DerKrebs [107]

Answer:

Part 1) $c=12.14\ units$

Part 2) $m\angle A=41.78^o$

Part 3) $m\angle B=74.22^o$

Step-by-step explanation:

step 1

Find the measure of side c

Applying the law of cosines

$c^2=a^2+b^2-2(a)(b)cos(C)$

we have

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substitute

$c^2=9^2+13^2-2(9)(13)cos(64^o)$

$c^2=81+169-234cos(64^o)$

$c^2=250-234cos(64^o)$

$c^2=147.4212$

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step 2

Find the measure of angle A

Applying the law of sines

$\frac{a}{sin(A)}=\frac{c}{sin(C)}$

substitute the given values

$\frac{9}{sin(A)}=\frac{12.14}{sin(64^o)}$

solve for sin(A)

$sin(A)=\frac{sin(64^o)}{12.14}(9)$

$sin(A)=0.6663$

$m\angle A=sin^{-1}(0.6663)=41.78^o$

step 3

Find the measure of angle B

we know that

The sum of the interior angles in any triangle must be equal to 180 degrees

so

$m\angle A+m\angle B+m\angle C=180^o$

substitute the given values

$41.78^o+m\angle B+64^o=180^o$

$m\angle B+105.78^o=180^o$

$m\angle B=180^o-105.78^o$

$m\angle B=74.22^o$

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