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Andrew [12]
3 years ago
8

What is the distance, to the nearest tenth

Mathematics
1 answer:
Alecsey [184]3 years ago
7 0

I think the answer may be B or c but I would say c

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Plz help i will mark brainliest
Pepsi [2]

Answer:

x = 9.

Step-by-step explanation:

42 - 15 = 27

27 ÷ 3 = 9

4 0
3 years ago
What is the value of 6/x+2x^2, when x=3
Elden [556K]

Answer:

20

Step-by-step explanation:

6/x + 2x²

6/3 + 2(3)²

2 + 2(9)

2 + 18

20

4 0
3 years ago
Christelle and Larkin both drive customers on the weekend. Christelle charges a flat fee of $4, plus $1 per mile. Larkin charges
ozzi

Complete question :

Christelle and Larkin both drive customers on the weekend. Christelle charges a flat fee of $4, plus $1 per mile. Larkin charges a flat fee of $2, plus $1.25 per mile.

Which equations represent each of their earnings, (y), after a given number of miles, (x)?

Select two that apply. Put more then one answer

y=4x+4

y=x+4

y=2x+1.25

y=1.25x+2

y=−x−4

y=1.25x−2

Answer:

y=x+4

y=1.25x + 2

Step-by-step explanation:

Christelle:

Flat fee = $4

Charge per mile = $1

Earning, y after x miles

y = 4 + x

Lakin:

Flat fee = $2

Charhe per mile = $1.25

EEarning, y after x miles

y = 2 + 1.25x

5 0
3 years ago
PLEASE HELP ME ASAP BECAUSE I AM SLOWLY LOSING MY MIND-
Blababa [14]

Answer:

35 + 80 = 115

180 - 115 = 65

plzz brainliest........

3 0
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.

3 0
3 years ago
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