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3 years ago

13

Calculate the slope of the line passing through the two points: (3,-3) & (1,-7)

1 answer:

DanielleElmas [232]3 years ago

8 0

Answer:

3×1=3 _3×7=9

Step-by-step explanation:

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This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the ex

olga nikolaevna [1]

$f(x_1,\ldots,x_n)=x_1+\cdots+x_n=\displaystyle\sum_{i=1}^nx_i$

${x_1}^2+\cdots+{x_n}^2=\displaystyle\sum_{i=1}^n{x_i}^2=4$

The Lagrangian is

$L(x_1,\ldots,x_n,\lambda)=\displaystyle\sum_{i=1}^nx_i+\lambda\left(\sum_{i=1}^n{x_i}^2-4\right)$

with partial derivatives (all set equal to 0)

$L_{x_i}=1+2\lambda x_i=0\implies x_i=-\dfrac1{2\lambda}$

for $1\le i\le n$ , and

$L_\lambda=\displaystyle\sum_{i=1}^n{x_i}^2-4=0$

Substituting each $x_i$ into the second sum gives

$\displaystyle\sum_{i=1}^n\left(-\frac1{2\lambda}\right)^2=4\implies\dfrac n{4\lambda^2}=4\implies\lambda=\pm\frac{\sqrt n}4$

Then we get two critical points,

$x_i=-\dfrac1{2\frac{\sqrt n}4}=-\dfrac2{\sqrt n}$

or

$x_i=-\dfrac1{2\left(-\frac{\sqrt n}4\right)}=\dfrac2{\sqrt n}$

At these points we get a value of $f(x_1,\cdots,x_n)=\pm2\sqrt n$ , i.e. a maximum value of $2\sqrt n$ and a minimum value of $-2\sqrt n$ .

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3 years ago

PLZ HELP DUE TODAY THE ANSWER CHOICES ARE

timama [110]

From the last time i did it, i believe its B

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3 years ago

X − 6 = 63 − 10(x + 8)

il63 [147K]

$\red { \green {\boxed {\boxed{Answer}}}}$

$x - 6 =63 - 10(x + 8)$

$x - 6 = 63 - 10x - 80$

$x - 6 = 63 - 80 - 10x$

$x - 6 = - 17 - 10x$

$x - 6 + 10x = - 17$

$x + 10 x- 6 = - 17$

$11x - 6 = - 17$

$11x = - 17 + 6$

$11x = - 11$

$x = \frac{ - 11}{11}$

$x = - 1$

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Evaluate the following: (Your answers should have an i in them)

Yuliya22 [10]

Answer:

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