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marissa [1.9K]
4 years ago
5

Sarah is purchasing pencils to share. Each package has 12 pencils. The equation n = 12p, where n is the total number of pencils

and p is the number of packages, can be used to determine the total number of pencils Sarah purchased. Determine which variable is dependent and which is independent. Then, make a table showing the number of pencils purchased for 3–7 packages.
Mathematics
1 answer:
podryga [215]4 years ago
7 0

Answer and Step-by-step explanation:

Each package has 12 pencils

total number of pencils Sarah purchased: n = 12p

n = total number of pencils

p = number of packages

n depends on p, because the total number of pencils dependos on how many packages she will buy, so:

independent variable: p

dependent variable: n

╔═══╦══════════╦════╗

║ p     ║ n = 12p            ║ n  ║

╠═══╬══════════╬════╣

║ 3     ║ n = 12*3           ║ 36     ║

╠═══╬══════════╬════╣

║ 4     ║ n = 12*4           ║ 48     ║

╠═══╬══════════╬════╣

║ 5     ║ n = 12*5           ║ 60     ║

╠═══╬══════════╬════╣

║ 6     ║ n = 12*6           ║ 72     ║

╠═══╬══════════╬════╣

║ 7     ║ n = 12*7           ║ 84     ║

╚═══╩══════════╩════╝

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leonid [27]

Answer:

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

(Please vote me Brainliest if this helped!)

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)
  • N\left(y\right)=y,\:\quad M\left(x\right)=x^3-3x+3

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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