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MrMuchimi
3 years ago
6

Factor Xsquared +12x+27

Mathematics
1 answer:
____ [38]3 years ago
3 0

Answer:

(x+3)(x+9)

Step-by-step explanation:

We are given the expression x^2+12x+27

This means that we are looking for two numbers that would add together to give us 12 and multiply to give us 27.

The two numbers that fit this description are 3 and 9.

This means that the factored form of the expression would be

(x+3)(x+9)

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Find Integrating Fector: <br> ylogy dx + ( x-logy )dy=0
12345 [234]
Rewrite the ODE as

y\log y\,\mathrm dx+(x-\log y)\,\mathrm dy=0\iff y\log y\dfrac{\mathrm dx}{\mathrm dy}+x=\log y
\dfrac{\mathrm dx}{\mathrm dy}=\dfrac1{y\log y}x=\dfrac1y

so that it is now linear in x. An integrating factor would

\mu(y)=\exp\left(\displaystyle\int\frac{\mathrm dy}{y\log y}\right)=e^{\log(\log y)}=\log y

Multiply both sides by \mu(y) to get

\log y\dfrac{\mathrm dx}{\mathrm dy}=\dfrac1yx=\dfrac{\log y}y
\dfrac{\mathrm d}{\mathrm dy}[x\log y]=\dfrac{\log y}y
x\log y=\displaystyle\int\frac{\log y}y\,\mathrm dy
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4 0
4 years ago
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Vikentia [17]
4 x 6 = 24.
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4 0
3 years ago
Solve this system of linear inequalities ( will give brainiest)
lakkis [162]
From the graphs you can see that the graph of the line y= \frac{1}{3} x-3 lies under the graph of the line y= \frac{1}{3} x-1. Then all solutions of unequality y\le \frac{1}{3} x-3 are solutions of unequality y\le \frac{1}{3} x-1, but not all solutions of unequality y\le \frac{1}{3} x-1 are solutions of unequality y\le \frac{1}{3} x-3. 
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4 years ago
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DENIUS [597]

I think the answer is A) 2.


I apologize if this is incorrect

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3 years ago
Simplify 3(x+2)-9x+5
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It would be...
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6 0
3 years ago
Read 2 more answers
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