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den301095 [7]
3 years ago
12

What is the length of the segment whose endpoints are A(-4, 3) and B (10, 6)?

Mathematics
2 answers:
nlexa [21]3 years ago
7 0
Apply Pythagoras:

length = sqrt( (10--4)² + (6-3)² ) = sqrt(205)
Alik [6]3 years ago
5 0

Answer:  The correct option is (B) \sqrt{205}.

Step-by-step explanation:  We are given to find the length of the segment whose endpoints are A(-4, 3) and B (10, 6).

We know that

the length of a line segment with endpoints (a, b) and (c, d) is calculated using distance formula as follows :

D=\sqrt{(c-a)^2+(d-b)^2}.

Therefore, using distance formula, the distance between the endpoints A(-4, 3) and B (10, 6) is given by

AB\\\\=\sqrt{(10-(-4))^2+(6-3)^2}\\\\=\sqrt{(10+4)^2+3^2}\\\\=\sqrt{14^2+9}\\\\=\sqrt{196+9}\\\\=\sqrt{205}.

Thus, the length of the line segment AB is \sqrt{205}~\textup{units}.

Option (B) is CORRECT.

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How do you know that 35x5=5x35 without finding the products?
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Here is the assignment : You are "selling" the polynomial. You must convince the buyer you have the Best polynomial on the marke
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Step-by-step explanation:

A polynomial is an expression that can be built from constants and symbols called variables or indeterminates by means of addition, multiplication and exponentiation to a non-negative integer power. Two such expressions that may be transformed, one to the other, by applying the usual properties of commutativity, associativity and distributivity of addition and multiplication, are considered as defining the same polynomial.

A polynomial in a single indeterminate x can always be written (or rewritten) in the form

{\displaystyle a_{n}x^{n}+a_{n-1}x^{n-1}+\dotsb +a_{2}x^{2}+a_{1}x+a_{0},}a_{n}x^{n}+a_{n-1}x^{n-1}+\dotsb +a_{2}x^{2}+a_{1}x+a_{0},

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This can be expressed more concisely by using summation notation:

{\displaystyle \sum _{k=0}^{n}a_{k}x^{k}}{\displaystyle \sum _{k=0}^{n}a_{k}x^{k}}

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3 0
3 years ago
Consider all 5 letter "words" made from the full English alphabet. (a) How many of these words are there total? (b) How many of
VARVARA [1.3K]

Answer:

a) There are 11,881,336 of these words in total.

b) There are 7,893,600 of these words with no repeated letters.

c) 896,376 of these words start with an a or end with a z or both

Step-by-step explanation:

Our words have the following format:

L1 - L2 - L3 - L4 - L5

In which L1 is the first letter, L2 the second letter, etc...

There are 26 letters in the English alphabet.

(a) How many of these words are there total?

Each of L1, L2, L3, L4 and L5 have 26 possible options.

So there are 26^{5} = 11,881,336 of these words total

(b) How many of these words contain no repeated letters?

The first letter can be any of them, so L1 = 26.

At the second letter, the first one cannot be repeated, so L2 = L1 - 1 = 25.

At the third letter, nor the first nor the second one can be repeated, so L3 = L1 - 2 = 24

This logic applies until L5

So we have

26-25-24-23-22

In total there are

26*25*24*23*22 = 7,893,600

of these words with no repeated letters.

(c) How many of these words start with an a or end with a z or both (repeated letters are allowed)?

T = T_{1} + T_{2} + T_{3}

T_{1} is the number of words that start with an a and do not end with z. So we have

1 - 26 - 26 - 26 - 25

The first letter can only be a, and the last one cannot be z. So:

T_{1} = 26^{3}*25 = 439,400

T_{2} is the number of words that start with any letter other than a and end with z. So we have

25 - 26 - 26 - 25 - 1

The first letter can be any of them, other than a, and the last can only be z. So:

T_{2} = 26^{3}*25 = 439,400

T_{3} is the number of words that both start with a and end with z. So:

1 - 26 - 26 - 26 - 1

The first letter can only be a, and the last can only be z. The other three letters could be anything. So:

T_{3} = 26^{3} = 17,576

T = T_{1} + T_{2} + T_{3} = 2*439,400 + 17,576 = 896,376

896,376 of these words start with an a or end with a z or both

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3 years ago
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