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2 years ago

3. Determine whether f(x) = -5x2- 10x + 6 has a maximum or a minimum

Mathematics

1 answer:

Sati [7]2 years ago

6 0

-2(2+5x) because -5x2-10x+6 = -10-10x+6=-4-10x which equals the answer which is -2(2+5x)

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2 years ago

Need answer please show work

Jet001 [13]

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1 year ago

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3 years ago

Solve the following equations x1/3=2 <br> Y-2=9<br> (2s)1/2=9<br> 2n-1=16

Nookie1986 [14]

$x \times \frac{1}{3} = 2 \\ = > x = 2 \times 3 \\ = > x = 6$

$y - 2 = 9 \\ = > y = 9 + 2 = 11$

$(2s) \frac{1}{2} = 9 \\ = > s = 9$

$2n - 1 = 16 \\ = > 2n = 16 + 1 \\ = > 2n = 17 \\ = > n = \frac{17}{2} = 8.5$

<h3>The answers are :</h3><h3>x = 6</h3><h3>y = 11</h3><h3>s = 9</h3><h3>n = 8.5</h3><h3>Hope it helps!</h3><h3 />

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2 years ago

Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.

fredd [130]

$\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

$\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

so the ODE is indeed exact and there is a solution of the form $F(x,y)=C$ . We have

$\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)$

$\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)$

$f'(y)=0\implies f(y)=C$

$\implies F(x,y)=x^2y^3+\ln(x+y^2)=C$

With $y(1)=2$ , we have

$8+\ln9=C$

$\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}$

8 0

2 years ago