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Blababa [14]
3 years ago
12

Solve for a. 5a - 2 = 23 a = ?

Mathematics
2 answers:
snow_tiger [21]3 years ago
3 0
5a - 2 = 23.

First, take 5a - 2 = 23 and add 2 to each side

5a - 2 = 23
     +2    +2
5a = 25

now, divide each side by 5

5a = 25
/5      /5

a = 5
kirza4 [7]3 years ago
3 0
Solution

1) <span>Add 2 to both sides
</span><span>5a=23+2
</span>
2) <span>Simplify <span>23+2</span> to 25
</span><span>5a=25
</span>
3) <span>Divide both sides by 5
</span>a= \frac{25}{5}

4)Simplify \frac{25}{5} <span>to 5
</span><span>a=5

ANSWER = 5</span>
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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}}

(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)
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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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