What type of function is represented in the table?

Sketch the lines given by x+y =6 and −3x+y = 2 on the same set of axes to solve the system graphically.Then solve the system of

tankabanditka [31]

Answer:

x= 1 and y= 5

The graphical solution is in the attachment.

Step-by-step explanation:

There's a lot of methods for solving a system of equations. For example, the substitution method.

You have to solve one of the equations for one variable (x or y) and replace it in the other equation and solve it for that variable. Therefore you will obtain a solution and then you have to replace that solution in the first equation in order to obtain the second solution.

In this case, solving the first equation for x:

x+y=6

Adding -y both sides:

x + y - y = 6-y

x = 6-y (I)

Replacing it in the second equation:

-3x+y = 2

-3(6-y) + y = 2

Applying the distributive property:

-3(6) -3(-y) + y = 2

-18 +3y + y= 2

Adding 18 both sides:

18 -18 +4y = 2 + 18

4y = 20

Dividing by 4 both sides:

y = 20/4 =5

Replacing it in (I)

x= 6 - 5

x = 1

In order to sketch the given lines, you have to obtain two points for each one and trace the line containing those points. The solution of the system of equations is the point of intersection of the two lines.

-The intersection with the x-axis is given by replacing y=0

x + 0 = 6

x=6

P1(6,0)

-The intersection with the y-axsis is given by replacing x=0

0 + y =6

y = 6

P2(0,6)

-The intersection with the x-axis:

-3x + 0 =2

x = -2/3

P3(-2/3,0)

-The intersection with the y-axis:

-3(0) + y = 2

y=2

P4(0,2)

The intersection of the two lines is setting the equations equal to each other:

y = 6-x and y = 3x + 2

6-x = 3x +2

Solving for x:

6 - 2 = 3x + x

4 = 4x

x = 1

And replacing x=1 in any of the equations you will obtain the solution point (1,5)

3 0

3 years ago

Suppose that the time to complete a test is bell-shaped distributed with mean 45 minutes and standard deviation 15 minutes. if t

lisov135 [29]

75-45/15= 2. 1-p-value of 2 (.9772)= .0228

5 0

3 years ago

My math problem dealing with fractions ask What is 1/8 of 9?

Norma-Jean [14]

Answer:

a

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Paige and Thuy are on a hill by a river. Paige throws a stone into the river. Her

Drupady [299]

The answer is Y=9x-1

7 0

4 years ago

Read 2 more answers

Proof by induction on the number of horses: Basis Step. There is only one horse. Then clearly all horses have the same color. In

Novosadov [1.4K]

Answer:

Claiming mathematical induction, of the statement: "all horses are the same color", the theorem is a counterfeit paradox sustained by mistaken demonstrations.

Step-by-step explanation:

”that is a horse of a different color” was a familiar expression in the middle of the last century, meaning that something is quite different from normal or common expectation, but George Polya, a great mathematician provided proof that there is no horse of a different color:

Theorem: "All horses are the same color"

Proof (by induction on the number of horses):

- Base Case: P(1) is undoubtedly true, as having only one horse, then all horses have the same color.

- Inductive Hypothesis: Assume P(n), which is the statement that n horses all have the same color.

- Inductive Step: Given a set of n+1 horses {h1,h2,...,hn+1}, we can eliminate the last horse in the serie and use the inductive hypothesis onlky to the first n horses {h1,...,hn}, deducing that they all have the same color. The same way, the conclusion may be that the last n horses {h2,...,hn+1} all have the same color. But the “middle” horses {h2,...,hn} (i.e., all but the first and the last) belong to both of these series, so they have the same color as horse h1 and horse hn+1. It follows, therefore, that all n+1 horses have the same color. Therefore, using the principle of induction, all horses have the same color.

It is clear that, it is not true that all horses are of the same color, so where is the mistake in our induction proof? It is tempting to blame the induction hypothesis. But even though the induction hypothesis is false (for n ≥ 2), that is not the mistaken reasoning. The real flaw in the proof is that the induction step is valid for a “typical” value of n, say, n = 3. The flaw, however, is in the induction step when n = 1. In this case, for n+1 = 2 horses, there are no “middle” horses, this makes the argument to collapse.

7 0

3 years ago