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Mathematics

2 answers:

IRISSAK [1]2 years ago

8 0

Answer:

1. y<-2 2. x≤5 3. 0<23 4. 0 ≤ -1 5. inequality: 3 + (7x) < 24 answer: x<3

Step-by-step explanation:

Ymorist [56]2 years ago

7 0

1. 7 - 6y > 9

7 - 6y > 9

Subtract 7 from both sides

-6y > 19 - 7

Subtract 7 from 19 to get 12

-6y > 12

Divide both sides by -6 (because the 6 is negative, the direction of the inequality is changed)

y < 12/-6

Divide 12 by -6 to get -2

y < -2 (graph this on the number line)

2. 2(x+2) - 3x ≥ -1

2 (x+2) - 3x ≥ -1

Use the distributive property to multiply 2 by x+2

2x + 4 -3x ≥ -1

Combine 2x and -3x to get -x

-x + 4 ≥ -1

Subtract 4 from both sides

-x ≥ -1 - 4

Subtract 4 from -1 to get -5

-x ≥ -5

Divide both sides by -1 (again because the 1 is negative, the direction of the inequality is changed)

x ≤ -5/-1

The fraction can be simplified to 5 by removing both negative signs

x ≤ 5 (graph this on the number line)

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Find three real numbers x, y, and z whose sum is 6 and the sum of whose squares is as small as possible. g

mart [117]

We're minimizing $x^2+y^2+z^2$ subject to $x+y+z=6$ . Using Lagrange multipliers, we have the Lagrangian

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x+y+z-6)$

with partial derivatives

$\begin{cases}L_x=2x+\lambda\\L_y=2y+\lambda\\L_z=2z+\lambda\\L_\lambda=x+y+z-6\end{cases}$

Set each partial derivative equal to 0:

$\begin{cases}2x+\lambda=0\\2y+\lambda=0\\2z+\lambda=0\\x+y+z=6\end{cases}$

Subtracting the second equation from the first, we find

$2x-2y=0\implies x=y$

Similarly, we can determine that $x=z$ and $y=z$ by taking any two of the first three equations. So if $x=y=z$ determines a critical point, then

$x+y+z=3x=6\implies x=y=z=2$

So the smallest value for the sum of squares is $2^2+2^2+2^2=12$ when $(x,y,z)=(2,2,2)$ .