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3 years ago

Least common multiple 14,6

Mathematics

1 answer:

Delicious77 [7]3 years ago

6 0

Answer:

Step-by-step explanation:

Multiples of 14 are:

0 , 14 , 28 , 42. ... .. .

Multiples of 6 are :

0, 6 , 12 , 18 , 24 , 30 , 36 , 42 ... ... . .

Another method is to write the prime factorization of 14 and 6:

Take all the prime numbers with the highest exponent from the prime factorization:

14 = 2 × 7 and 6 = 2 × 3

L . C . M (14 , 6 ) = 2 ×3 × 7 = 42

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Solve the inequality: -4x -7>1

anastassius [24]

Answer:

1. x < -2

2. Yes

Step-by-step explanation:

-4x - 7 > 1

= -4x > 8

= x < -2

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3 years ago

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Answer:

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B)6.5 calories

Step-by-step explanation:

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Step-by-step explanation:

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3 years ago

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Norma-Jean [14]

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3 years ago

Read 2 more answers

Exercise 3.7.4: let a = 2 1 0 0 2 0 0 0 2 .

swat32

With

$\mathbf A=\begin{bmatrix}2&1&0\\0&2&0\\0&0&2\end{bmatrix}$

we have

$\det(\mathbf A-\lambda\mathbf I)=\begin{vmatrix}2-\lambda&1&0\\0&2-\lambda&0\\0&0&2-\lambda\end{vmatrix}=(2-\lambda)^3$

so $\mathbf A$ has one eigenvalue, $\lambda=2$ , with multiplicity 3.

In order for $\mathbf A$ to not be defective, we need the dimension of the eigenspace to match the multiplicity of the repeated eigenvalue 2. But $\mathbf A-2\mathbf I$ has nullspace of dimension 2, since

$\begin{bmatrix}0&1&0\\0&0&0\\0&0&0\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\mathbf 0\implies x=0\text{ or }y=0$

That is, we can only obtain 2 eigenvectors,

$\begin{bmatrix}1\\0\\0\end{bmatrix}\text{ and }\begin{bmatrix}0\\0\\1\end{bmatrix}$

and there is no other. We needed 3 in order to complete the basis of eigenvectors.

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3 years ago