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nasty-shy [4]
2 years ago
14

If g(x)=(x-3) exponent 2, what is the value of g(-9)

Mathematics
2 answers:
sleet_krkn [62]2 years ago
6 0

Answer:

144

Step-by-step explanation:

g(-9) is found by substituting for x = -9 in the function g(x) = (x-3)²

We get g(-9) = (-9-3)² = (-12)² = 144

kenny6666 [7]2 years ago
3 0

Step-by-step explanation:

g(x) = (x - 3)²

g(-9) = (-9 - 3)²

g(-9) = (-12)²

g(-9) = 144

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π·6^2

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Solve for w.<br><br> w/5 = 3<br> 0<br> 2<br> 3<br> 5<br> 8<br> 15<br> 
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Step-by-step explanation:

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6 0
3 years ago
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HELP PLEASEE FOR HW!!!
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Please please please help and solve this with steps, much help needed thank you :) 20 points for this!
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}}

7 0
3 years ago
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