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forsale [732]
3 years ago
15

Pls help extra points and mark brainlist

Mathematics
1 answer:
igor_vitrenko [27]3 years ago
5 0

Answer:

2over/a -b-c

Step-by-step explanation:

remove the parentheses

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What is the perimeter of rhombus LMNO? 20 units 24 units 40 units 48 units
NeTakaya
The little lines on each side of the rhombus mean that all the sides are the same length.

We can set line LM and MN equal to solve for X, then we can solve the length of a side.

3x-3 = x+7

Add 3 to each side:
3x = x +10

Subtract x from each side:

2x = 10

Divide both sides by 2:
x = 10/2
x = 5

Now we have the value for x, replace x in one of the side formulas:

x +7 = 5+7 = 12

Each side = 12 units.

 The perimeter would be 12 + 12 + 12 + 12 = 48 units.



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3 years ago
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Complete the missing parts of the paragraph proof.
mariarad [96]
THANK YOU FOR YOUR HELP.
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What is the least common multiple of the numbers 5,25, and 15
liubo4ka [24]

Answer:

5

Step-by-step explanation:

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3 years ago
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What is x+y=10;y=x+10
Alenkasestr [34]

[ Answer ]

\boxed{\bold{X \ = \ 0, \ Y \ = \ 10}}

[ Explanation ]

  • System Of Equations \bold{\left \{ {X \ + \ Y \ = \ 10} \atop {Y \ + \ X \ = \ 10}} \right.}

-----------------------------------

  • [Substitute] Y = X + 10

\bold{\begin{bmatrix}x+x+10=10\end{bmatrix}}

  • Isolate x for x + x + 10 = 10: x = 0

For y = 10

Substitute x = 0

Y = 0 + 10

0 + 10 = 10

y = 10

  • Solutions

X = 0

Y = 10

\boxed{\bold{[] \ Eclipsed \ []}}

3 0
3 years ago
Inverse laplace of [(1/s^2)-(48/s^5)]
Katen [24]
**Refresh page if you see [ tex ]**

I am not familiar with Laplace transforms, so my explanation probably won't help, but given that for two Laplace transform F(s) and G(s), then \mathcal{L}^{-1}\{aF(s)+bG(s)\} = a\mathcal{L}^{-1}\{F(s)\}+b\mathcal{L}^{-1}\{G(s)\}

Given that \dfrac{1}{s^2} = \dfrac{1!}{s^2} and -\dfrac{48}{s^5} = -2\cdot\dfrac{4!}{s^5}

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2} - 2\cdot\dfrac{4!}{s^5}\right\} = \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\}

From Table of Laplace Transform, you have \mathcal{L}\{t^n\} = \dfrac{n!}{s^{n+1}} and hence \mathcal{L}^{-1}\left\{\dfrac{n!}{s^{n+1}}\right\} = t^n

So you have \mathcal{L}^{-1}\left\{\dfrac{1}{s^2}\right\} - 2\mathcal{L}^{-1}\left\{\dfrac{4!}{s^5}\right\} = \boxed{t-2t^4}.

Hope this helps...
7 0
3 years ago
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