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3 years ago

I need help :( this problem is very confusing

Mathematics

2 answers:

Sladkaya [172]3 years ago

6 0

Answer:

Emily , don't put your whole name in the questions :DDDD, it's at the top of the page, anyway, Eloper .. :DDD

Step-by-step explanation:

It's arithmetic b/c you're just adding 1

arithmetic

are the answers , the last one is b/c that's the common difference.

Ksju [112]3 years ago

3 0

Arithmetic sequences are when a sequence of number adds or subtracts the same number every time to the following number. 1,2,3,4 is just 1+1=2 2+1=3. Therefore, it is arithmetic.

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FromTheMoon [43]

I believe 198 or 197 cars were in the Garage

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3 years ago

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What is the correct interpretation of the point (250, 0)? The average price per ticket is $250. The minimum price per ticket is

Brut [27]

The point (250,0) of the graph represents that the average price per ticket is $250.

Given to us

x is the price the passenger paid

f(x) is the positive percent difference

<h3>What is the correct interpretation of the point (250, 0)?</h3>

We know that a coordinate is written in the form of (x, y), therefore, the point (250, 0) represents that the price of the ticket is 250, while the 0 in the coordinate represents that there is no percentage difference. Since the point (250,0) is the mid-value of the x-axis on the graph, we can say that $250 is the average price of the ticket.

Hence, the point (250,0) of the graph represents that the average price per ticket is $250.

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2 years ago

what IS THE FUTURE VALUE OF THE 10% savings from earnings of 36,000 if it earns 6.25% annual interest compounded quarterly for 1

bezimeni [28]

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4 0

3 years ago

Which fraction would make this equation true 5×?=6 1/4

Blizzard [7]

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3 years ago

Find the inverse of ????−7 −2

Masja [62]

Answer:

$A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

Step-by-step explanation:

$A=\begin{bmatrix}-7 & -2\\ 4 & 1\end{bmatrix}$

$det\ A=-7\times 1-(-2)\times 4\\\Rightarrow det\ A=1$

$A^{-1}=\frac{1}{det\ A}\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A^{-1}=\frac{1}{1}\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

$\therefore A^{-1}=\begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}$

$A.A^{-1}=\begin{bmatrix}-7 & -2\\ 4 & 1\end{bmatrix}\times \begin{bmatrix}1 & 2\\ -4 & -7\end{bmatrix}\\\Rightarrow A.A^{-1}=\begin{pmatrix}\left(-7\right)\cdot \:1+\left(-2\right)\left(-4\right)&\left(-7\right)\cdot \:2+\left(-2\right)\left(-7\right)\\ 4\cdot \:1+1\cdot \left(-4\right)&4\cdot \:2+1\cdot \left(-7\right)\end{pmatrix}\\\Rightarrow A.A^{-1}=\begin{pmatrix}1&0\\ 0&1\end{pmatrix}$

Hence, confirmed

6 0

3 years ago