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eimsori [14]
3 years ago
14

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

equations algebraically to verify your graphical solution.

Mathematics
1 answer:
tankabanditka [31]3 years ago
3 0

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.

<u>For x+y = 6</u>

-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)

<u>For -3x + y = 2</u>

-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)

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As you were in a hurry to catch a flight, you picked two out of your four credit cards at random, without even looking. One of t
agasfer [191]

Answer:

P = 0.5

Step-by-step explanation:

There are four cards.

Lets say,

Card 1 = A

Card 2 = B

Card 3 = C

Card 4 = D

Lets suppose that A has reached its limit. Listing all possibilities of picking 2 random cards.

1st possibility = A and B

2nd possibility = A and C

3rd possibility = A and D

4th possibility = B and C

5th possibility = B and D

6th possibility = C and D

There are three possibilities that the card which has reached its limit will be picked with the usable card. Calculation probability.

P = total number of wanted outcomes / total number of possible outcomes

P = 3/6

P = 0.5

3 0
3 years ago
A sample survey contacted an SRS of 2854 registered voters shortly before the 2012 presidential election and asked respondents w
Brilliant_brown [7]

Answer:

a) 97.37%

b) 1.31%

Step-by-step explanation:

a)  

Here we want to calculate the area under the Normal curve with mean 0.52 and standard deviation 0.009 between 0.5 and 0.54

This can be easily done with a spreadsheet and we get

<em>P (0.5くV < 0.54) = 0.9737 or 97.37% </em>

(See picture 1)

b)

Here we want the area under the Normal curve with mean 0.52 and standard deviation 0.009 to the left of 0.5.

<em>P(V ≤ 0.5) = 0.0131 or 1.31% </em>

(See picture 2)

3 0
3 years ago
Write the first three terms of the geometric sequence that has a1= 120 and whose common ratio is r = -1/2
Alona [7]

For geometric sequence, you multiply the common ratio to the previous one.


120 × (-1/2) = -60

-60 × (-1/2) = 30

30 × (-1/2) = -15


So your first three terms is

-60 , 30 , -15


Hope this helped! Brainliest is always welcome :)

8 0
4 years ago
A quarter horse often excels at running races of 1/4 mile or less .how many feet is that
romanna [79]

Answer:

1320ft

Step-by-step explanation:

1 mile is equal to 5280 feet, divide 5280/4

this will result in a quarter mile, being 1320ft. So for the question "1/4 mile or less" the Answer is the horse excels in 1320ft mile races

3 0
3 years ago
A rectangle is constructed with its base on the diameter of a semicircle with radius 2 and with its two other vertices on the se
stealth61 [152]

Answer:

L = 2*√2

w = √2

Step-by-step explanation:

Given:

A rectangle is constructed with its base on the diameter of a semicircle with radius 2 and with its two other vertices on the semicircle.

Find:

What are the dimensions of the rectangle with maximum​ area?

Solution:

- Let the length and width of the rectangle be L and w respectively.

- We know that Length L lie on the diameter base. So , L < 4 and the width w is less than 2 . w < 2.

- Using the Pythagorean Theorem, we relate the L with w using the radius r = 2 of the semicircle.

                           r^2 = (L/2)^2 + (w)^2

                           sqrt (4 - w^2 ) = L / 2

                           L = 2*sqrt (4 - w^2 )           L < 4 , w < 2

- The relation derived above is the constraint equation and the function is Area A which is function of both L and w as follows:

                          A ( L , w ) = L*w

- We substitute the constraint into our function A:

                          A ( w ) = 2*w*sqrt (4 - w^2 )

- Now we will find the critical points for width w for which A'(w) = 0

                         A'(w) = 2*sqrt (4 - w^2 ) - 2*w^2 / sqrt (4 - w^2 )  

                         0 = [2*sqrt (4 - w^2 )*sqrt (4 - w^2 )   - 2*w^2] / sqrt (4 - w^2 )  

                         0 = 2*(4 - w^2 )   - 2*w^2

                         0 = -4*w^2 + 8

                         8/4 = w^2

                         w = + sqrt ( 2 )   ..... 0 < w < 2

- From constraint equation we have:

                          L = 2*sqrt (4 - 2 )

                          L = 2*sqrt(2)

7 0
4 years ago
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