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3 years ago

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A science teacher told his students that the expression 4m^4/144n^4 represents the area in square units of the contaminated squa

re region in their activity. What is the width of the region in simplest form?

2 answers:

Bond [772]3 years ago

8 0

Given that the contaminated region is a square, the area is given by the formula:
Area=length*width
since in a square length=width
the
area=length×length
or
area=width×width

thus given that area=4m^4/144n^4 sq. units let the width be w units
thus
area=w×w=4m^4/144n^4
w²=4m^4/144n^4
solving for w we get:
w=√(4m^4/144n^4 )
w=2m²/12n²
Hence the width,w=2m²/12n²

Anon25 [30]3 years ago

8 0

Answer:

The answer is 100% <u>B. m^2/6n^2 units</u>

I took the test and it was correct :)

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If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

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$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

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