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mrs_skeptik [129]
3 years ago
13

SOMEONE PLEASE HELP....

Mathematics
1 answer:
Anton [14]3 years ago
6 0

Answer:

Step-by-step explanation:

(3x+4)(x^2+px+5) \\=3x^3+3px^2+15x+4x^2+4px+20\\=3x^3+(3p+4)x^2+(15+4p)x +20

since we know that x^2  in this expansion is -23x^{2}

thus 3p+4=-23\\3p=-27\\p=-9

Hope that helps!

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Answer:

3:4:7

Step-by-step explanation:

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Can someone PLZZZZ HELP!!!!
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Answer:

The solution is:

Part A. \sqrt{5}^{\frac{7k}{3}}) which is sqrt(5)^7k/3[/tex]

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Step-by-step explanation:

Part A.

To solve this part, we're going two use THREE important properties of exponents:

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Let's work the numerator using the properties 1, 2 and 3:

(\sqrt{5}^{3} )^{\frac{k}{9} } }  = (\sqrt{5}^{3\frac{k}{9}}) = (\sqrt{5}^{\frac{k}{3}})

Let's work the denominator using the properties 1, 2 and 3:

(\sqrt{5}^{6} )^{-\frac{k}{3} } }  = (\sqrt{5}^{6\frac{k}{3}}) = (\sqrt{5}^(2k))

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\sqrt{5}^{\frac{k}{3}-(-2k)})=\sqrt{5}^{\frac{7k}{3}})

Part B

if 5^{\frac{3}{2} } 5^{\frac{3}{2}} = \sqrt{5}^{\frac{7k}{3}})

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5^{3} = 5^{\frac{7k}{6}})

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k = 18/7

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kobusy [5.1K]
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