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3 years ago

252 = 42 x 6

Mathematics

1 answer:

max2010maxim [7]3 years ago

3 0

I think it’s D but it not to sure ☺️

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1. Identify the LCM for 6 and 18. O 36 O 6 O 18 O 108

Sphinxa [80]

Answer: 18

Step-by-step explanation: 6 x 3 = 18, 18 x 1 = 18

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3 years ago

Which shape have at least one right angle choose are that are correct

arsen [322]

Answer:

Square and rectangle

The other polygon isn't necessarily have right angle

7 0

3 years ago

Read 2 more answers

Which values of a, b, and c represent the answer in simplest form?

Hoochie [10]

The value of a is 1 , b = 3 , c = 4 , Option B is the correct answer.

The correct question is

Which values of a, b, and c represent the answer in simplest form?

7/9 divided by 4/9 = $\rm a\dfrac{b}{c}$

1: a = 1, b = 4, c = 3

2: a = 1, b = 3, c = 4

3: a = 1, b = 63, c = 36

4: a = 1, b = 36, c = 63

<h3>What is reducing to simplest Form ?</h3>

Reducing to simplest form means to reduce to an extent where the numerator and denominator have nothing in common to be cancelled out.

It is given in the question that

(7/9) / (4/9) = $\rm a\dfrac{b}{c}$

It can also be written as

(7*9)/(4*9) = $\rm a\dfrac{b}{c}$

7/4 = $\rm a\dfrac{b}{c}$

$\rm 1 \dfrac{3}{4} = a\dfrac{b}{c}$

Therefore the value of a is 1 , b = 3 , c = 4 , Option B is the correct answer.

To know more about Simplest form

brainly.com/question/290068

#SPJ1

3 0

2 years ago

Let an = –3an-1 + 10an-2 with initial conditions a1 = 29 and a2 = –47. a) Write the first 5 terms of the recurrence relation. b)

zlopas [31]

We can express the recurrence,

$\begin{cases}a_1=29\\a_2=-47\\a_n=-3a_{n-1}+10a_{n-2}7\text{for }n\ge3\end{cases}$

in matrix form as

$\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}$

By substitution,

$\begin{bmatrix}a_{n-1}\\a_{n-2}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}\begin{bmatrix}a_{n-2}\\a_{n-3}\end{bmatrix}\implies\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}^2\begin{bmatrix}a_{n-2}\\a_{n-3}\end{bmatrix}$

and continuing in this way we would find that

$\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}\begin{bmatrix}a_2\\a_1\end{bmatrix}$

Diagonalizing the coefficient matrix gives us

$\begin{bmatrix}-3&10\\1&0\end{bmatrix}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}-5&0\\0&2\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}$

which makes taking the $(n-2)$ -th power trivial:

$\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}-5&0\\0&2\end{bmatrix}^{n-2}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}$

$\begin{bmatrix}-3&10\\1&0\end{bmatrix}^{n-2}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}(-5)^{n-2}&0\\0&2^{n-2}\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}$

So we have

$\begin{bmatrix}a_n\\a_{n-1}\end{bmatrix}=\begin{bmatrix}-5&2\\1&1\end{bmatrix}\begin{bmatrix}(-5)^{n-2}&0\\0&2^{n-2}\end{bmatrix}\begin{bmatrix}-5&2\\1&1\end{bmatrix}^{-1}\begin{bmatrix}a_2\\a_1\end{bmatrix}$