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tester [92]
2 years ago
15

Select the graph for the solution of the open sentence Ix| = -2

Mathematics
1 answer:
Oliga [24]2 years ago
7 0

Answer:

Attached graph

Step-by-step explanation:

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x=722/69

Step-by-step explanation:

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3 years ago
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Step-by-step explanation:

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7 0
3 years ago
Solve the equation for x.
Alona [7]

the answer would be 41/5 or 8 1/5

4 0
3 years ago
You have 24 months left until you graduate and you plan on buying yourself a new $20,000 car on graduation day. If you invest $3
shusha [124]

Answer: No, the money won't be enough to buy the car

Step-by-step explanation:

you plan on buying yourself a new $20,000 car on graduation day and graduation day is 24 months time. If you invest $300 a month for the next 24 months.

The principal amount, p = 300

He is earning 4% a month, it means that it was compounded once in four months. This also means that it was compounded quarterly. So

n = 4

The rate at which the principal was compounded is 4%. So

r = 4/100 = 0.04

It was compounded for a total of 24 months. This is equivalent to 2 years. So

n = 2

The formula for compound interest is

A = P(1+r/n)^nt

A = total amount that would be compounded at the end of n years.

A = 300(1 + (0.04/4)/4)^4×2

A = 300(1 + 0.01)^8

A = 300(1.01)^8

A = $324.857

The total amount at the end of 24 months is below the cost of the car which is $20000. So he won't have enough money to buy the car

3 0
3 years ago
Given the matrices: 1 2 A= 1 -1 2 1 1 B= 3 4 Calculate AB: C11 C12 [2.1] х 1 2 3 4 C21 C22 C11 = C12 = -2 C22 - C215 DONE​
eduard

Answer:

\:c_{11}=-2,\:\:\:c_{12}=-2

\:c_{21}=5,\:\:\:c_{22}=8

Step-by-step explanation:

Given the matrices

A=\begin{pmatrix}1&-1\\ 2&1\end{pmatrix}

B=\begin{pmatrix}1&2\\ \:3&4\end{pmatrix}

Calculating AB:

\begin{pmatrix}1&-1\\ \:\:2&1\end{pmatrix}\times \:\begin{pmatrix}1&2\\ \:\:3&4\end{pmatrix}=\begin{pmatrix}c_{11}&c_{12}\\ \:\:\:c_{21}&c_{22}\end{pmatrix}

Multiply the rows of the first matrix by the columns of the second matrix

                                   =\begin{pmatrix}1\cdot \:1+\left(-1\right)\cdot \:3&1\cdot \:2+\left(-1\right)\cdot \:4\\ 2\cdot \:1+1\cdot \:3&2\cdot \:2+1\cdot \:4\end{pmatrix}

                                   =\begin{pmatrix}-2&-2\\ 5&8\end{pmatrix}

Hence,

\begin{pmatrix}c_{11}&c_{12}\\ \:\:\:c_{21}&c_{22}\end{pmatrix}=\begin{pmatrix}-2&-2\\ \:5&8\end{pmatrix}

Therefore,

\:c_{11}=-2,\:\:\:c_{12}=-2

\:c_{21}=5,\:\:\:c_{22}=8

5 0
3 years ago
Read 2 more answers
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