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Oliga [24]
3 years ago
5

How do u simplify 15x^4 y^2 over 40x^3 y^2 

Mathematics
1 answer:
Vinil7 [7]3 years ago
4 0
\frac{15x^4y^2}{40x^3y^2}=\frac{3}{8}x^{4-3}y^{2-2}=\frac{3}{8}x^1y^0=\frac{3}{8}x
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Which of the following is a true statement?
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It's D. That is the only true statement.
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Find g(4x)<br> g(x)=x²-4
7nadin3 [17]

Answer:

g(x) = x² - 4 is already in form of a variable, I.e., x

g(4x) takes another variable, I.e., 4x

Same as before, 4x takes over x:

=> g(4x) = (4x)² - 4

  • <em>(</em><em>ax</em><em>)</em><em>²</em><em> </em><em>=</em><em> </em><em>a</em><em>²</em><em>x</em><em>²</em><em>,</em><em> </em><em>where</em><em> </em><em>a</em><em> </em><em>is</em><em> </em><em>some</em><em> </em><em>arbitrary</em><em> </em><em>constant</em><em>.</em><em> </em>

<h3><u>Answer</u><u>:</u> </h3>

=> g(4x) = 16x² - 4

OR

=> g(4x) = 4{4x² - 1}

3 0
2 years ago
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Help me! <br>This is Algebra problem
aleksklad [387]

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3 years ago
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alukav5142 [94]
The characteristic solution follows from solving the characteristic equation,

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which has second derivative

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Substituting into the ODE, you have

y''+4y=\cos2x+\sin2x
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y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x

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but, of course, any solution of the form a_0\cos2x+b_0\sin2x is already accounted for within y_c.
6 0
2 years ago
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