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2 years ago

8

250 divided by 10 squared equals?

2 answers:

navik [9.2K]2 years ago

4 0

250 divided by 10 squared is 2.5

Anna35 [415]2 years ago

3 0

$\frac{250}{10^2}$
$= \frac{250}{100}$
$= 2.5$

In short, the answer would be : 2.5.

Hope this helps !

Photon

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Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

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.

ii. The Lagrangian is

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

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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2 years ago

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Softa [21]

yeah i dont understand what that even means.

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3 years ago

Use f(x)=(2x+1)/(3x-2) and g(x)=5x-1 to find <br> (gof)(x)

Orlov [11]

Answer:

From the functions f(x) and g(x) we have $(g\circ f)(x)=\frac{7(x+1)}{3x-2}$

Step-by-step explanation:

Given that the functions f is defined by $f(x)=\frac{2x+1}{3x-2}$

and g is $g(x)=5x-1$

To find the $(g\circ f)(x)$ :

$(g\circ f)(x)=g(f(x))$

$=g(\frac{2x+1}{3x-2})$

$=5(\frac{2x+1}{3x-2})-1$

$=\frac{10x+5}{3x-2}-1$

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$=\frac{10x+5-3x+2}{3x-2}$

$=\frac{7x+7}{3x-2}$ ( adding the like terms )

$=\frac{7(x+1)}{3x-2}$

Therefore $(g\circ f)(x)=\frac{7(x+1)}{3x-2}$ .

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If there are 4000 cars stuck in line due to a traffic accident and each car is 15.84 feet in length, how many miles long is the

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12 miles

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Answer-
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