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4 years ago

9

25 POINTS ---need an explanation

1 answer:

olganol [36]4 years ago

7 0

I suggest you look at the work in the pictures I attached. It examplains how you work out the problem and what the answer is

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How much does an ice cream cost with 11 ounces of toppings?

oksano4ka [1.4K]

<em>$3.00 for ice cream </em>

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<em>+ $2.00 for the 4 ounces of toppings </em>

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<em>$5.00 </em>

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<em>hope it helps!!</em>

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3 years ago

Please help ASAP!<br> Thank you

Leno4ka [110]

The answer has to be 16 because when you add them it goes up

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3 years ago

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Points P and Q are two of the vertices of a right triangle. They are the endpoints of the hypotenuse PQ¯¯¯¯¯ of the triangle. Th

kolezko [41]

Answer:

.

Step-by-step explanation:

7 0

2 years ago

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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.

5 0

3 years ago

Help (10 points ) ;)

joja [24]

Answer:

c

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers