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marissa [1.9K]
3 years ago
8

When 6a^2 b^3• -3a^5 b^3 multiplied, the product of the coefficients is .

Mathematics
2 answers:
ad-work [718]3 years ago
7 0

Answer: last option

Step-by-step explanation:

To solve this problem you must apply the properties of the exponents. WHen you have two powers with equal base, you must add the exponents.

Then you have:

(6a^{2}b^{3})(-3a^{5}b^{3})=-18a^{(2+5)}b^{(3+3)}=-18a^{7}b^{6}

Therefore as you can see, the coefficient is -18

Vlad1618 [11]3 years ago
5 0

Answer:

-18

Step-by-step explanation:

Question says to multiply 6a^2 b^3 and -3a^5 b^3.

to find the product of the coefficients. And select correct choice from given list:

3

-3

18

-18

Coefficient of 6a^2 b^3 is 6

Coefficient of -3a^5 b^3 is -3

Then product of coefficients = (6)(-3)=-18

Hence -18 is correct choice.

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Answer:

y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}

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Step-by-step explanation:

\frac{dy}{dx}y=x^3+2x-5x+3

\mathrm{First\:order\:separable\:Ordinary\:Differential\:Equation}

  • \mathrm{A\:first\:order\:separable\:ODE\:has\:the\:form\:of}\:N\left(y\right)\cdot y'=M\left(x\right)

1. \mathrm{Substitute\quad }\frac{dy}{dx}\mathrm{\:with\:}y'\:

y'\:y=x^3+2x-5x+3

2. \mathrm{Rewrite\:in\:the\:form\:of\:a\:first\:order\:separable\:ODE}

yy'\:=x^3-3x+3

  • N\left(y\right)\cdot y'\:=M\left(x\right)
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3. \mathrm{Solve\:}\:yy'\:=x^3-3x+3:\quad \frac{y^2}{2}=\frac{x^4}{4}-\frac{3x^2}{2}+3x+c_1

4. \mathrm{Isolate}\:y:\quad y=\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+4c_1}{2}}

5. \mathrm{Simplify}

y=\sqrt{\frac{x^4-6x^2+12x+c_1}{2}},\:y=-\sqrt{\frac{x^4-6x^2+12x+c_1}{2}}

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