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2 years ago

12

May someone please help me with 1-5? I have been stuck on this for a while now.

1 answer:

AleksAgata [21]2 years ago

5 0

Answer:

1 and 2 are correct!! Good Job!!

3. on the number line put a solid dot over 0, the line should to the right.

4. on the number line put a solid dot over -4, the line will go to the left.

5. on the number line put a open dot over 1.5, the line will go to the left.

Step-by-step explanation:

I hope this helps!! Also, you did great on 1 and 2, got this!!

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(1,000,000 centimeters) x (1 skewer/30centimeters) =

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2 years ago

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14. Find the Error Hiroshi is determining -42 – (-37). Find his

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Answer:

Hiroshi's error was not having changed the sign of the number 37 by removing the parentheses and placing -37 instead of +37.

Step-by-step explanation:

we have

Hiroshi's work

$42-(-37)=42-37=5$

Remember that

a negative number is just a positive number multiplied by -1

and (-1)(-1)=+1

so

Correct work

$42-(-37)$

$42-(-1)(37)$

$42-(1)(-1)(37)$

$42+(-1)(-1)(37)$

$=42+37=79$

therefore

Hiroshi's error was not having changed the sign of the number 37 by removing the parentheses and placing -37 instead of +37.

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2 years ago

Find the absolute maximum and minimum values of f(x,y)=xy−4x in the region bounded by the x-axis and the parabola y=16−x2.

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Answer:

The absolute maximum and minimum is $20\; \text{and} -20.$

Step-by-step explanation:

We first check the critical points on the interior of the domain using the

first derivative test.

$f_x=y-4=0$

$f_y=x=0$

The only solution to this system of equations is the point (0, 4), which lies in the domain.

$f_{xx}=0, \;f_{yy}=0\; \text{and}\; f_{xy}=-1$

$\Rightarrow f_{xx}f_{yy}-f_{xy}=o-1=-1$

$\therefore (0,4)$ is a saddle point.

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Along boundary $y=16-x^2$

$f=x(16-x^2)-4x$

$=16x-x^3-4x$

$\Rightarrow f^'=16-3x^2-4=0$

$\Rightarrow 3x^2=12$

$\Rightarrow x=\pm2,\;\;y=14$

Values of f(x) at these points.

$\begin{array}{}(4,0)=-16\\(-4,0)=16\\(0,16)=0\\(2,14)=20\\(-2,14)=-20\end{array}$

Therefore, the absolute maximum and minimum is $20\; \text{and} -20.$

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