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3 years ago

Which statement about the following system of inequalities is true?

Mathematics

1 answer:

elena-14-01-66 [18.8K]3 years ago

3 0

Answer:

the following system of inequalities is true

Step-by-step explanation:

there is no solution because the graph do not intersect

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PLEASE HELP PLEASE!!!

geniusboy [140]

Answer:

$48.30

Step-by-step explanation:

Total profit after selling 5 cases:

$5 \times 19.50 = 97.50$

Net loss:

$145.80 - 97.50 = 48.30$

8 0

4 years ago

Read 2 more answers

Solve by factoring x^2 + 8x =9

Helen [10]

Answer:

$x=-9,\:x=1$

Step-by-step explanation:

$x^2+8x=9\\x^2+8x-9=0\\(x+9)(x-1)=0\\x=-9,\:x=1$

6 0

2 years ago

Carlo attends art class every 4 weeks, chess club every 2 weeks, and fencing every 3 weeks. If he attended all three this week,

Aneli [31]

In 12 weeks as the lowest common number between them is 12
4+4+4=12
3+3+3+3=12
2+2+2+2+2+2=12

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4 years ago

Rachel read 30 minutes for every 10 minutes that she spent watching television. Nicolas read 45 minutes for every 15 minutes tha

stich3 [128]

Answer:

B.different

Step-by-step explanation:

Rachel read 30 minutes for every 10 minutes that she spent watching television

30 - 10 = 20

Nicolas read 45 minutes for every 15 minutes that he watched television

45 - 15 = 30

Hope It Helps Friend :)

4 0

3 years ago

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.

3 0

3 years ago