All categories

3 years ago

X+y+z=3. 2x-2y+4z=12. 6x+2y+3z=-11

Mathematics

2 answers:

Sedaia [141]3 years ago

7 0

X = -3
y = 3
z = 3
this is the answer i got idk if that is what you were looking for

Ymorist [56]3 years ago

4 0

Use elimination with the bottom two equations to get 8x + 7z = 1

Multiply the first equation by 2 and then do the same thing with the first two equations.

4x + 6z = 18

Now with these two new equations that only have two variables, do elimination again after multiplying 4x + 6z = 18 by -2 so that the x terms cancel out.

-5z = -35

z = 7

Now plug 7 in for z in one of the new equations to find x.

4x + 6(7) = 18

4x = -24

x = -6

Now use one of the original equations and plug 7 and -6 in to find y.

-6 + y + 7 = 3

y = 2

You might be interested in

what is the distance between point S and U? Round to the nearest tenth of a unit? A 3.3 B 6.4 C 8.1 D 12.0

ki77a [65]

The distance between the points (d) is found using the Pythagorean theorem.

Imagine the two points as defining the hypotenuse of a right triangle. The lengths of the legs of the triangle are the horizontal distance between the points and the vertical distance between the points. The the theorem tells us
d² = 4² + 7²
d² = 16 + 49
d² = 65
d = √65 ≈ 8.1

The distance between the points is
C 8.1

_____
You know that the distance must be longer than the longest leg (7) and must be shorter than the sum of the two legs (4+7=11). The only answer choice between 7 and 11 is 8.1.

8 0

3 years ago

BRAINLIEST AND 20 POINTS

Korvikt [17]

Tan(60)=Opposite or L /20cm

*20. *20

Tan(60)*20=L

34.64101615=L

34.64 (to the nearest hundredth of a meter or cm)

5 0

3 years ago

Read 2 more answers

The answer to the problem

swat32

.bahajaiakijajwjwjwjwjwjnwjwq

5 0

3 years ago

In the game of roulette, a player can place a $9 bet on the number 11, and have a 1/38 probability of winning. If the metal ba

damaskus [11]

Answer:

1. What is the expected value of the game to the player?

Expected value= -$0.47368

2. If you played the game 1000 times, how much would you expect to lose?

Expected value= -$473.68

Step-by-step explanation:

1. What is the expected value of the game to the player?

Expected value can tell how much you will gain/lose every time you do the game. To find the expected value you need to multiply every value with its chance to occur. There are two types of events here, winning and losing. You have 1/38 chance of winning $315 and 37/38 chance of losing $9.

The expected value will be:

expected value= 1/38 * $315 + 37/38*(-$9) = $8.2894 - $8.7631

expected value= -$0.47368

You are expected to lose $0.47368 every time you play

2. If you played the game 1000 times, how much would you expect to lose?

Expected value calculated for each roll. If you want to know how much you gain/lose by multiple rolls, you just need to multiply the expected value with the number of rolls. The game has negative expected value so you are expected to lose money instead of gaining money. The money you expected to lose will be:

number of roll * expected value

1000 timer * -$0.47368/time= -$473.68

You are expected to lose $473.68 if you play the game 1000 times.

6 0

3 years ago

Suppose a goat was tethered at the middle of a 100-foot-long fence and the length of the rope was 50 feet, as shown in the diagr

Dvinal [7]

The area the goat can graze is 1/2π(50)²= 1250π sq. feet

The rope length is 50 feet again, thus the results are the same.

1/2π(50)²= 1250π sq. feet

8 0

3 years ago