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2 years ago

The following equations are given:

1 answer:

3 0

The solution to the variables using equations 1, 2 and 3 is x = 2, y = -1 and z = 3 and the solutions in part b hold for equation 4

<h3>Solving the variables using equations 1 and 2</h3>

There are three variables in the system

To solve the three variables, there must be at least three equations in the system of equations

Hence, the variables cannot be solved because the number of equations is not enough

<h3>Solving the variables using equations 1, 2 and 3</h3>

We have:

3x + z + y = 8

5y - x = -7

3z + 2x - 2y = 15

Make x the subject in (2)

x = 5y + 7

Substitute x = 5y + 7 in (1) and (3)

3(5y + 7) + z + y = 8

15y + 21 + z + y = 8

16y + z = -13

3z + 2(5y + 7) - 2y = 15

3z + 10y + 14 - 2y = 15

3z + 8y = 1

Multiply 16y + z = -13 by 3

48y + 3z = -39

Subtract 3z + 8y = 1 from 48y + 3z = -39

48y - 8y + 3z - 3z = -39 - 1

40y = -40

Divide by 40

y = -1

Substitute y = -1 in x = 5y + 7

x = 5(-1) + 7

x = 2

Make z the subject in 16y + z = -13

z = -13 - 16y

Substitute y = -1

z = -13 - 16(-1)

z = 3

Hence, the solution to the variables using equations 1, 2 and 3 is x = 2, y = -1 and z = 3

<h3>Can the solution work for equation 4?</h3>

We have:

4x + 5y - 2z = -3

Substitute x = 2, y = -1 and z = 3

4(2) + 5(-1) - 2(3) = -3

Evaluate

-3 = -3 --- this is true

Hence, the solutions in part b hold for equation 4

Read more about system of equations at:

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Missing part of the question

The following equations are given:

Equation #1: 3x + z + y = 8

Equation #2: 5y - x = -7

Equation #3: 3z + 2x - 2y = 15

Equation #4: 4x + 5y - 2z = -3

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Divide. x2 + 2x - 49 X +8 The answer is (Simplify your answer.)

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Answer:

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Step-by-step explanation:

x2+2x−49X+8

2−49+2+8

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3 years ago

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Can you guys pls help me with this

insens350 [35]

1.probability of not rolling a factor 5 =25/36

2.probability of rolling a difference of 3=1/6

3.probability of rolling the product of 12=1/9

4.probability of rolling the product 6=1/9

5.probability of rolling factors of 2 on second die=1/3

6.probability of not rolling multiples of 4 on both die=25/36

7.probability of not rolling a sum of 5=8/9

8.probability of rolling a difference 1=10/36

9.probability of not rolling factors of 6 on both die=4/6

10.probability of not rolling prime numbers on both dice=1/4

Step-by-step explanation:

To solve all possible probabilities, we have to find out the sample space of the experiment of rolling two dice simultaneously.

Sample space S={(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)}

{(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)}

{(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)}

{(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)}

{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)}

{(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

probability of not rolling a factor 5

possible set={(1,1),...(1,4),(1,6),(2,1),....(2,4),(2,6),(3,1)......(3,4),(3,6),(4,1)...(4,4),(4,6),(6,1)....(6,4),(6,6)}

number of events n(E)=25

p(not a factor of 5)=n(E)/n(S)=25/36

probability of rolling a difference of 3

E={(1,4),(2,5),(3,6),(4,1),(5,2),(6,3)}

number of events=6

P(difference of 3)=n(E)/n(S)=6/36=1/6

probability of rolling the product of 12

E={(2,6),(3,4),(4,3),(6,2)}

n(E)=4

P(E)=n(E)/n(S)=4/36=1/9

probability of rolling the product 6

E={(1,6),(2,3),(3,2),(6,1)}

n(E)=4

P(E)=n(E)/n(S)=4/36=1/9

probability of rolling factors of 2 on second die

E={(1,1),(1,2),(2,1),(2,2),(3,1),(3,2),(4,1),(4,2),(5,1),(5,2),(6,1),(6,2)}

n(E)=12

P(E)=n(E)/n(S)=12/36=1/3

probability of not rolling multiples of 4 on both die

E={(1,1),(1,2),(1,3),(1,5),(1,6),(2,1),(2,2),(2,3),(2,5),(2,6),(3,1),(3,2),(3,3),(3,5),(3,6),(5,1),(5,2),(5,3),(5,5),(5,6),(6,1),(6,2),(6,3),(6,5),(6,6)}

n(E)=25

P(E)=n(E)/n(S)=25/36

probability of not rolling a sum of 5

E={(1,1),(1,2),(1,3),(1,5),(1,6),(2,1),(2,2),(2,4),(2,5),(2,6),(3,1),(3,3),(3,4),(3,5),(3,6),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),......(5,6),(6,1),......(6,6)

n(E)=32

P(E)=n(E)/n(S)=32/36=8/9

probability of rolling a difference 1

E={(1,2),(2,1),(2,3),(3,2),(3,4),(4,3),(4,5),(5,4),(5,6),(6,5)}

n(E)=10

P(E)=n(E)/n(S)=10/36

probability of not rolling factors of 6 on both die

E={(4,4),(4,5),(5,4),(5,5)}

n(E)=4

P(E)=n(E)/n(S)=4/6

probability of rolling prime numbers on both dice

E={(2,2),(2,3),(2,5),(3,2),(3,3),(3,5),(5,2),(5,3),(5,5)}

n(E)=9

P(E)=n(E)/n(S)=9/36=1/4

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I have a circular necklace with $18$ beads on it. All the beads are different. Making two cuts with a pair of scissors, I can di

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Solution:

There are four general types we can make: (6,12),(7,11),(8,10),(9,9).

First type: (6,12), there are 18 possible ways to choose those 6, which are going to be cut from the necklace: choose a direction to count the beads, and choose a starting position (between 2 beads). There are exactly 18 starting positions, since there are 18 spaces between the beads. One cut is equivalent for a pair we can make.

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Third type: (8,10), with the same reasoning, there are 18 possible ways to cut the necklace.

Fourth type: (9,9), The same reasoning cannot be applied again, since half of the cuts would be exactly the same as the other half. So there are 9 possible cuts, exactly one or each axis of symmetry for the necklace.

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3 years ago

Tammy rents an apartment close to her school campus. The amount that she spends on rent is given by the equation r = 355m, where

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Tammy rents an apartment close to her school campus. The equation that is given is this one:
r = 355m,
where r is the amount spend on the rent.
m is the number of months.

In order to get the constant proportionality of r to m in this relationship, then we can set a constant k variable.

r = 355 m (k)
k = 1 r / 355 m

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4 0

3 years ago

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