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marin [14]
3 years ago
5

Assume that a chocolate bar consists of n squares arranged in a rectangular pattern. The entire bar, a smaller rectangularpieceo

fthebar,canbebrokenalongavertical orahorizontallineseparatingthesquares.Assumingthat only one piece can be broken at a time, determine how manybreaksyoumustsuccessivelymaketobreakthebar into n separate squares. Use strong induction to prove your answer.
Mathematics
1 answer:
motikmotik3 years ago
8 0

Step-by-step explanation:

Claim:

it takes n - 1 number of breaks to break the bar into n separate squares for all integers n.

Basic case -> n = 1

The bar is already completely broken into pieces.

Case -> n ≥ 2

Assuming that assertion is true for all rectangular bars with fewer than n squares. Break the bar into two pieces of size k and n - k where 1 ≤ k < n

The bar  with k squares requires k − 1 breaks and the bar with n − k squares

requires n − k − 1 breaks.

So the original bar requires  1 + (k−1) + (n−k−1) breaks.

simplifying yields,

1 + k − 1 + n − k − 1

1 - 1 + n - 1

n - 1

Therefore, we proved as we claimed that it takes n - 1 breaks to break the bar into n separate squares.

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Lerok [7]

Answer:

LN corresponds to RQ

Step-by-step explanation:

If we look at the diagram we find that LN corresponds to RQ, MN corresponds to QP and ML corresponds to PR .

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Therefore

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3 years ago
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Varvara68 [4.7K]

Answer:

Explanation and problem 1 solved below.

Step-by-step explanation:

The quadrants are numbered this way:

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Quadrant 2: upper left

Quadrant 3: lower left

Quadrant 4: lower right

Signs of coordinates in the quadrants:

x-coordinate is positive in Q1 and Q4

x-coordinate is negative in Q2 and Q3

y-coordinate is positive in Q1 and Q2

y-coordinate is negative in Q3 and Q4

1. (-1, 5)

x is negative, so it is Q2 or Q3

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Answer: quadrant 2

4 0
3 years ago
Kite DCFE is inscribed in circle A shown below.
kogti [31]

Answer:

∠DEF =56°

Step-by-step explanation:

We are given that arc DEF = 248°

So, arc DCF = 360°-248°

arc DCF = 112°

Now we are required to find the ∠DEF.

So, we will use Inscribed angle theorem.  

Inscribed angle theorem: The measure of an inscribed angle is equal to one-half the measure of its intercepted arc.  

Intercepted arc = arc DCF = 112°

Inscribed angle = ∠DEF

⇒∠DEF = \frac{1}{2}arcDCF

⇒∠DEF = \frac{1}{2}\times 112^{\circ}

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let's bear in mind that Y is a bisecting point, so it's really cutting XZ into two equal halves.

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