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3 years ago

PLEASE HELP ME IT HAS BEEN A VERY ROUGH MORNING THANK YOU!

Mathematics

1 answer:

IgorLugansk [536]3 years ago

3 0

Answer: D

Step-by-step explanation:

10/2=5

15/3=5

20/4=5

15/5=3

D doesn’t belong.

Hope this helps :)

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Question 7 A number cube with faces labeled from 1 to 6 will be rolled once. The number rolled will be recorded as the outcome.

QveST [7]

Answer:

1) $S=\{1,2,3,4,5,6\}$

2) $O=\{1,3,5\}$

Step-by-step explanation:

Given : A number cube with faces labeled from 1 to 6 will be rolled once. The number rolled will be recorded as the outcome.

To find : Give the sample space describing all possible outcomes. Then give all of the outcomes for the event of rolling an odd number. If there is more than one element in the set, separate them with commas ?

Solution :

A number cube with faces labeled from 1 to 6 will be rolled once.

The sample space describing all possible outcomes is given by

$S=\{1,2,3,4,5,6\}$

All of the outcomes for the event of rolling an odd number is given by

$O=\{1,3,5\}$

6 0

3 years ago

Solve the formula for the perimeter of a rectangle, P = 21 + 2w for l

Phoenix [80]

2(w+l)=21+2w

w+l=10.5+w

So l = 10.5

7 0

3 years ago

Find all points on the portion of the plane x+y+z=5 in the first octant at which f(x, y, z) = xy2z2 has a maximum value.

irina1246 [14]

Lagrange multipliers:

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

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

$\lambda=-y^2z^2=-2xyz^2=-2xy^2z$

$-y^2z^2=-2xyz^2\implies y=2x$ (if $y,z\neq0$ )

$-y^2z^2=-2xy^2z\implies z=2x$ (if $y,z\neq0$ )

$-2xyz^2=-2xy^2z\implies z=y$ (if $x,y,z\neq0$ )

In the first octant, we assume $x,y,z>0$ , so we can ignore the caveats above. Now,

$x+y+z=5\iff x+2x+2x=5x=5\implies x=1\implies y=z=2$

so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of $f(1,2,2)=16$ .

We also need to check the boundary of the region, i.e. the intersection of $x+y+z=5$ with the three coordinate axes. But in each case, we would end up setting at least one of the variables to 0, which would force $f(x,y,z)=0$ , so the point we found is the only extremum.