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3 years ago

Which coordinate pair is in the solution set for y < 1 − 6x?

1 answer:

4 0

As an open ended question there are infinitely many answers.
Graph the line y=-6x+1 and look for an answer that is below/to the left of the line.

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In how many ways can three married couples attending a concert be seated in a row of six seats if

OleMash [197]

i think it would be 30 ways

3 0

3 years ago

PlEaSe HeLp GiViNg BrAiNlIeSt

Romashka-Z-Leto [24]

2. The total number of ski trails is 12. That's how many numbers there are in the box.

3. The difference in length between the longest ski trail and the shortest ski trail is the biggest number minus the smallest number in the box. 3 1/8 minus 1 3/8 is 1 and 3/4. So the answer is 1.75 or 1 3/4, whichever one works.

4. There are 3 trails that are 2 7/8 miles long. so You do 2 7/8 times three or 2 7/8 plus 2 7/8 plus 2 7/8. That equals 8 19/40 which is the answer. Ti add them together first turn them into mixed numbers. 2 7/8 is the same as 23/8. 8x3=24. 23x3=69. 69/24 is the same is 8 19/40 through further simplification.

5. 3 1/8 plus 1 3/8 is 4 1/2 or 4.5

6. The shortest ski trail is 1 3/8. 1 3/8 times 3 is 4.125. 4 is greater than 3 which means Sam is automatically right.

8 0

3 years ago

Read 2 more answers

You have 16 digs for your current volleyball season. There are 3 games left in the season. You want to break your previous recor

natulia [17]

Answer: 16+d>20

d>4

Step-by-step explanation:

d=digs

16+d>20

16-16+d>20-16

d>4

7 0

2 years ago

Find the maximum and minimum values attained by f(x, y, z) = 5xyz on the unit ball x2 + y2 + z2 ≤ 1.

Allushta [10]

Check for critical points within the unit ball by solving for when the first-order partial derivatives vanish:
$f_x=5yz=0\implies y=0\text{ or }z=0$
$f_y=5xz=0\implies x=0\text{ or }z=0$
$f_z=5xy=0\implies x=0\text{ or }y=0$

Taken together, we find that (0, 0, 0) appears to be the only critical point on $f$ within the ball. At this point, we have $f(0,0,0)=0$ .

Now let's use the method of Lagrange multipliers to look for critical points on the boundary. We have the Lagrangian

$L(x,y,z,\lambda)=5xyz+\lambda(x^2+y^2+z^2-1)$

with partial derivatives (set to 0)

$L_x=5yz+2\lambda x=0$
$L_y=5xz+2\lambda y=0$
$L_z=5xy+2\lambda z=0$
$L_\lambda=x^2+y^2+z^2-1=0$

We then observe that

$xL_x+yL_y+zL_z=0\implies15xyz+2\lambda=0\implies\lambda=-\dfrac{15xyz}2$

So, ignoring the critical point we've already found at (0, 0, 0),

$5yz+2\left(-\dfrac{15xyz}2\right)x=0\implies5yz(1-3x^2)=0\implies x=\pm\dfrac1{\sqrt3}$
$5xz+2\left(-\dfrac{15xyz}2\right)y=0\implies5xz(1-3y^2)=0\implies y=\pm\dfrac1{\sqrt3}$
$5xy+2\left(-\dfrac{15xyz}2\right)z=0\implies5xy(1-3z^2)=0\implies z=\pm\dfrac1{\sqrt3}$

So ultimately, we have 9 critical points - 1 at the origin (0, 0, 0), and 8 at the various combinations of $\left(\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3},\pm\dfrac1{\sqrt3}\right)$ , at which points we get a value of either of $\pm\dfrac5{\sqrt3}$ , with the maximum being the positive value and the minimum being the negative one.

5 0

3 years ago

Which system of equations graphed below has a solution of (2,1)

goldfiish [28.3K]

Answer:

Sorry to say this... but neither of them have a solution of (2,1)

Step-by-step explanation:

The top one has a solution of (-2,-1). The one on the bottom has a solution of (-2,1). So... I'm not sure how this can help... but yeah...

8 0

3 years ago