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lakkis [162]
3 years ago
11

20 points.

Mathematics
1 answer:
attashe74 [19]3 years ago
3 0

Answer: 315

Step-by-step explanation:

375-25(coupon)=350

350-35(discount)=315

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zhannawk [14.2K]
-2=3x+0 Would Be The Equation
8 0
4 years ago
Pls help I really need it
levacccp [35]

Answer:

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Step-by-step explanation:

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7 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
Can one fifth be simplified
Svetradugi [14.3K]
No, and never, cus there ain't any numbers that are greater than 1 that can go into both of them.
<span>so basically if the numerator is 1, you can't simplify it anymore. Is Impossible.</span><span />
3 0
3 years ago
Helppppppppppp please
Ostrovityanka [42]

Answer:

72

Step-by-step explanation:

If we use the Intersecting chords theorem we can calculate that lengths AC and BD are equal calculating to 144.

So the diameter is 144,

Radius is half of the diameter.

144/2=72.

Hope this helped!

3 0
3 years ago
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