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3 years ago

B^2 - c^2 -10(b-c) Factor Completely people are getting this wrong I need help please

Mathematics

1 answer:

Slav-nsk [51]3 years ago

5 0

The factor that is the first one is the most important

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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,

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

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

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 $-2\sqrt n$ .

6 0

4 years ago

-15+7.5(2d-1)=7.5 what is d

snow_tiger [21]

Answer:

d=2

Step-by-step explanation:

3 0

4 years ago

Solve the system of equations.

nadya68 [22]

Answer:

using the first two equations cancelling the z

-2x-2y-5z=-7

6x+8y+5z=9

4x +6y= 2

using the second and the third equations cancelling the z

6x+8y+5z= 9

-2x-3y-5z= -6

4x +5y = 3

4x + 6y = 2

-4x -5y = -3

y= -1

2x + 2(-1) + 5z = 7

2x - 2 + 5z = 7

2x + 5z = 9

6x + 8(-1) + 5z = 9

6x - 8 + 5z = 9

6x + 5z = 17

-2x - 5z= -9

4x = 8

x = 2

2(2) + 2(-1) + 5z= 7

4 - 2 + 5z = 7

2 + 5z = 7

5z = 5

z = 1

(2, -1, 1)

the answer is a.

Step-by-step explanation:

3 0

4 years ago

How to do this question plz answer my question

ahrayia [7]

I gotchuuuu

Ok so angle ABC we’re gonna call it X

Angle BCD is 2X

The trick is the sum of angles of a pentagon is 540 degrees.

90 + 115 + 125 + 2X + X = 540
3X = 540-330
X = 70

SoOoOoOoOooooo

Angle BCD is 140 degrees !!

7 0

3 years ago

BRAINLIEST ASAP! PLEASE HELP ME :)

jenyasd209 [6]

<h3>Answers:</h3><h3>There are four solutions and they are</h3>

$\theta = 0$
$\theta = \pi$
$\theta = \frac{7\pi}{6}$ ... this says "7pi over 6"
$\theta = \frac{11\pi}{6}$ ... this says "11pi over 6"

===========================================

Work Shown:

$\sin(\theta)+1 = \cos(2\theta)\\\\\sin(\theta)+1 = 1-2\sin^2(\theta)\\\\\sin(\theta)+1-1+2\sin^2(\theta)=0\\\\2\sin^2(\theta)+\sin(\theta)=0\\\\\sin(\theta)(2\sin(\theta)+1)=0\\\\\sin(\theta)=0 \ \text{ or } \ 2\sin(\theta)+1=0\\\\$

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Solving $\sin(\theta)=0$ leads to $\theta=n\pi$ for any integer n.

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Solving $2\sin(\theta)+1=0$ leads to...

$2\sin(\theta)+1=0\\\\2\sin(\theta)=-1\\\\\sin(\theta)=-\frac{1}{2}\\\\\theta=\arcsin\left(-\frac{1}{2}\right) \ \text{ or } \ \theta=\pi-\arcsin\left(-\frac{1}{2}\right)\\\\\theta=\frac{11}{6}\pi+2\pi n \ \text{ or } \ \theta=\pi-\frac{11}{6}\pi+2\pi n\\\\\theta=\frac{11}{6}\pi+2\pi n \ \text{ or } \ \theta=\frac{-5}{6}\pi+2\pi n\\\\\theta=\frac{11}{6}\pi+2\pi n \ \text{ or } \ \theta=\frac{7}{6}\pi+2\pi n\\\\$

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The general solution set is

$\{\theta=n*\pi, \theta=\frac{11}{6}\pi+2\pi n, \theta=\frac{7}{6}\pi+2\pi n\}$

Again, n is any integer.

Let's look at a table of values where we plug in various integers for n. See the attached image below. Note the stuff in the highlighted yellow cells represents expressions that are between 0 and 2pi = 6.28

Therefore the four solutions are $\{0, \pi, \frac{7\pi}{6}, \frac{11\pi}{6}\}$ after we plug the proper values of n into the expressions, to have things match what the table shows.

6 0

3 years ago