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Rainbow [258]
3 years ago
14

Wondering if anyone could help?

Mathematics
1 answer:
aleksklad [387]3 years ago
4 0

Answer:

error

Step-by-step explanation:

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What is the value of expression?
kolezko [41]
6^{-7}\cdot6^4=6^{-7+4}=6^{-3}
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3 years ago
When does a sequence and series become an arithmetic sequence and arithmetic series​?
maria [59]

Answer:

An arithmetic sequence is a sequence of numbers in which the difference between the consecutive terms is constant.

4 0
2 years ago
jessie has a peice of wood that is 8 feet long he needs to cut pieces that are 7/8 of a foot long how many pieces will he be abl
attashe74 [19]

Answer: 9 pieces

Explanation: Divide 8 by 7/8 to find the number of cut pieces and you'll get 9.





3 0
3 years ago
Need help please anyone
Dafna11 [192]

Answer:

70°

Step-by-step explanation:

The minor angle of AB forms a straight line with angle x. When summed up it equals 180°.

minor angle of AB + x° = 180°

110° + x° = 180°

x° = 180° - 110°

= 70°

Enjoy!!!

5 0
3 years ago
Please solve the problem ​
jek_recluse [69]

Treat the matrices on the right side of each equation like you would a constant.

Let 2<em>X</em> + <em>Y</em> = <em>A</em> and 3<em>X</em> - 4<em>Y</em> = <em>B</em>.

Then you can eliminate <em>Y</em> by taking the sum

4<em>A</em> + <em>B</em> = 4 (2<em>X</em> + <em>Y</em>) + (3<em>X</em> - 4<em>Y</em>) = 11<em>X</em>

==>   <em>X</em> = (4<em>A</em> + <em>B</em>)/11

Similarly, you can eliminate <em>X</em> by using

-3<em>A</em> + 2<em>B</em> = -3 (2<em>X</em> + <em>Y</em>) + 2 (3<em>X</em> - 4<em>Y</em>) = -11<em>Y</em>

==>   <em>Y</em> = (3<em>A</em> - 2<em>B</em>)/11

It follows that

X=\dfrac4{11}\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\dfrac1{11}\begin{bmatrix}7&-10\\-7&11\end{bmatrix} \\\\ X=\dfrac1{11}\left(4\begin{bmatrix}12&-3\\10&22\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\left(\begin{bmatrix}48&-12\\40&88\end{bmatrix}+\begin{bmatrix}7&-10\\-7&11\end{bmatrix}\right) \\\\ X=\dfrac1{11}\begin{bmatrix}55&-22\\33&99\end{bmatrix} \\\\ X=\begin{bmatrix}5&-2\\3&9\end{bmatrix}

Similarly, you would find

Y=\begin{bmatrix}2&1\\4&4\end{bmatrix}

You can solve the second system in the same fashion. You would end up with

P=\begin{bmatrix}2&-3\\0&1\end{bmatrix} \text{ and } Q=\begin{bmatrix}1&2\\3&-1\end{bmatrix}

3 0
3 years ago
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