1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Vanyuwa [196]
3 years ago
11

Solve the linear system by using Substitution. Enter your answer as an ordered pair.

Mathematics
2 answers:
MAXImum [283]3 years ago
7 0

Answer: {y,x} = {4,2} ) ) ) )4

Step-by-step explanation: y

[2] y = -2x + 8

// Plug this in for variable y in equation [1]

[1] (-2x+8) - x = 2

[1] - 3x = -6

// Solve equation [1] for the variable x

[1] 3x = 6

[1] x = 2

// By now we know this much :

y = -2x+8

x = 2

// Use the x value to solve for y

y = -2(2)+8 = 4

Solution :

{y,x} = {4,2}

HACTEHA [7]3 years ago
6 0
2x+1x+2=8
3x+2=8
-2 -2
3x=6
divide 3 on both sides
x= 2
You might be interested in
It takes Bruce 12 h to complete his training. It takes Oliver approximately 2 days, 2 h, and 15 min to complete
Stolb23 [73]

Answer:

Oliver completed his training work in 50.25 hours

Step-by-step explanation:

we know that

1 day= 24 hours

1 hour=60 minutes

we have

2 days, 2 h, and 15 min

<u><em>Convert days to hours</em></u>

2 days=2(24)=48 hours

<u><em>Convert minutes to hour</em></u>

15 min=15(1/60)=1/4=0.25 h

so

2 days, 2 h, and 15 min=(48)+(2)+(0.25)=50.25 h

therefore

Oliver completed his training work in 50.25 hours

3 0
3 years ago
If an independent-measures design is being used to compare two treatment conditions, then how many different groups of participa
Serhud [2]

Answer:

2 groups and 1 score for each participant

Step-by-step explanation:

An independent measures design is a defined as a research method whereby multiple experimental groups are examined and the participants will only be in one group. Now, each participant will only be affected by one condition of the independent variable during the experiment.

In the question given, we want to use the method I just described to compare two treatment conditions.

This means that there will be two groups and each participant in both groups will be assigned one score.

7 0
3 years ago
Which doesnt belong, and why? 9, 16, 25, 43
Tomtit [17]

Answer:

43

Step-by-step explanation:

The first three numbers are perfect squares:

9 = 3²

16 = 4²

25 = 5²

43 is not a perfect square.

3 0
3 years ago
Read 2 more answers
Find a solution of x dy dx = y2 − y that passes through the indicated points. (a) (0, 1) y = (b) (0, 0) y = (c) 1 6 , 1 6 y = (d
Leni [432]
Answers: 

(a) y = \frac{1}{1 - Cx}, for any constant C

(b) Solution does not exist

(c) y = \frac{256}{256 - 15x}

(d) y = \frac{64}{64 - 15x}

Explanations:

(a) To solve the differential equation in the problem, we need to manipulate the equation such that the expression that involves y is on the left side of the equation and the expression that involves x is on the right side equation.

Note that

 x\frac{dy}{dx} = y^2 - y&#10;\\&#10;\\ \indent xdy = \left ( y^2 - y \right )dx&#10;\\&#10;\\ \indent \frac{dy}{y^2 - y} = \frac{dx}{x}&#10;\\&#10;\\ \indent \int {\frac{dy}{y^2 - y}} = \int {\frac{dx}{x}} &#10;\\&#10;\\ \indent \boxed{\int {\frac{dy}{y^2 - y}} = \ln x + C_1}      (1)

Now, we need to evaluate the indefinite integral on the left side of equation (1). Note that the denominator y² - y = y(y - 1). So, the denominator can be written as product of two polynomials. In this case, we can solve the indefinite integral using partial fractions.

Using partial fractions:

\frac{1}{y^2 - y} = \frac{1}{y(y - 1)} = \frac{A}{y - 1} + \frac{B}{y}&#10;\\&#10;\\ \indent \Rightarrow \frac{1}{y^2 - y} = \frac{Ay + B(y-1)}{y(y - 1)} &#10;\\&#10;\\ \indent \Rightarrow \boxed{\frac{1}{y^2 - y} = \frac{(A+B)y - B}{y^2 - y} }      (2)

Since equation (2) has the same denominator, the numerator has to be equal. So,

1 = (A+B)y - B&#10;\\&#10;\\ \indent \Rightarrow (A+B)y - B = 0y + 1&#10;\\&#10;\\ \indent \Rightarrow \begin{cases}&#10; A + B = 0&#10;& \text{(3)}\\-B = 1&#10; & \text{(4)}   \end{cases}

Based on equation (4), B = -1. By replacing this value to equation (3), we have

A + B = 0
A + (-1) = 0
A + (-1) + 1 = 0 + 1
A = 1 

Hence, 

\frac{1}{y^2 - y} = \frac{1}{y - 1} - \frac{1}{y}

So,

\int {\frac{dy}{y^2 - y}} = \int {\frac{dy}{y - 1}} - \int {\frac{dy}{y}} &#10;\\&#10;\\ \indent \indent \indent \indent = \ln (y-1) - \ln y&#10;\\&#10;\\ \indent  \boxed{\int {\frac{dy}{y^2 - y}} = \ln \left ( \frac{y-1}{y} \right ) + C_2}

Now, equation (1) becomes

\ln \left ( \frac{y-1}{y} \right ) + C_2 = \ln x + C_1&#10;\\&#10;\\ \indent \ln \left ( \frac{y-1}{y} \right ) = \ln x + C_1 - C_2&#10;\\&#10;\\ \indent  \frac{y-1}{y} = e^{C_1 - C_2}x&#10;\\&#10;\\ \indent  \frac{y-1}{y} = Cx, \text{ where } C = e^{C_1 - C_2}&#10;\\&#10;\\ \indent  1 - \frac{1}{y} = Cx&#10;\\&#10;\\ \indent \frac{1}{y} = 1 - Cx&#10;\\&#10;\\ \indent \boxed{y = \frac{1}{1 - Cx}}&#10;       (5)

At point (0, 1), x = 0, y = 1. Replacing these values in (5), we have

y = \frac{1}{1 - Cx}&#10;\\&#10;\\ \indent 1 = \frac{1}{1 - C(0)} = \frac{1}{1 - 0} = 1&#10;&#10;

Hence, for any constant C, the following solution will pass thru (0, 1):

\boxed{y = \frac{1}{1 - Cx}}

(b) Using equation (5) in problem (a),

y = \frac{1}{1 - Cx}   (6)

for any constant C.

Note that equation (6) is called the general solution. So, we just replace values of x and y in the equation and solve for constant C.

At point (0,0), x = 0, y =0. Then, we replace these values in equation (6) so that 

y = \frac{1}{1 - Cx}&#10;\\&#10;\\ \indent 0 = \frac{1}{1 - C(0)} = \frac{1}{1 - 0} = 1

Note that 0 = 1 is false. Hence, for any constant C, the solution that passes thru (0,0) does not exist.

(c) We use equation (6) in problem (b) and because equation (6) is the general solution, we just need to plug in the value of x and y to the equation and solve for constant C. 

At point (16, 16), x = 16, y = 16 and by replacing these values to the general solution, we have

y = \frac{1}{1 - Cx}&#10;\\&#10;\\ \indent 16 = \frac{1}{1 - C(16)} &#10;\\ &#10;\\ \indent 16 = \frac{1}{1 - 16C}&#10;\\&#10;\\ \indent 16(1 - 16C) = 1&#10;\\ \indent 16 - 256C = 1&#10;\\ \indent - 256C = -15&#10;\\ \indent \boxed{C = \frac{15}{256}}&#10;&#10;&#10;

By replacing this value of C, the general solution becomes

y = \frac{1}{1 - Cx}&#10;\\&#10;\\ \indent y = \frac{1}{1 - \frac{15}{256}x} &#10;\\ &#10;\\ \indent y = \frac{1}{\frac{256 - 15x}{256}}&#10;\\&#10;\\&#10;\\ \indent \boxed{y = \frac{256}{256 - 15x}}&#10;&#10;&#10;&#10;

This solution passes thru (16,16).

(d) We do the following steps that we did in problem (c):
        - Substitute the values of x and y to the general solution.
        - Solve for constant C

At point (4, 16), x = 4, y = 16. First, we replace x and y using these values so that 

y = \frac{1}{1 - Cx} &#10;\\ &#10;\\ \indent 16 = \frac{1}{1 - C(4)} &#10;\\ &#10;\\ \indent 16 = \frac{1}{1 - 4C} &#10;\\ &#10;\\ \indent 16(1 - 4C) = 1 &#10;\\ \indent 16 - 64C = 1 &#10;\\ \indent - 64C = -15 &#10;\\ \indent \boxed{C = \frac{15}{64}}

Now, we replace C using the derived value in the general solution. Then,

y = \frac{1}{1 - Cx} \\ \\ \indent y = \frac{1}{1 - \frac{15}{64}x} \\ \\ \indent y = \frac{1}{\frac{64 - 15x}{64}} \\ \\ \\ \indent \boxed{y = \frac{64}{64 - 15x}}
5 0
3 years ago
Please help! If you do tysm!!
AleksAgata [21]

Answer:

32 days

Step-by-step explanation:

5 divided by 2 is 2.5, which means 2.5 gallons will leak per day. 80 divided by 2.5 is 32 which means it will take 32 days for the bucket to completely leak out.

hope that's right <3

7 0
3 years ago
Other questions:
  • HELP PLEASE!
    8·2 answers
  • . The AutoCorrect feature can automatically capitalize the first letter in the names of days.
    5·1 answer
  • 2 Cereal comes in two different-sized boxes.
    8·1 answer
  • HELP HELP HELP PLEASE
    7·2 answers
  • Aleta’s puppy gained 3/8 pound each week for 4weeks . Although, how much weight did the puppy gain during 4 weeks
    14·1 answer
  • The population of city A was approximately 243,000 and it increased by 8% in one year.what was the new population
    10·1 answer
  • Use the perfect square trinomial identity to write this expression in factored form:
    15·2 answers
  • B. 80 = 30 =<br> Your answer
    5·1 answer
  • Tasha wants to find the quotient of 1026 divided by 27. She creates
    8·1 answer
  • Please help<br><br>answer choices are:<br><br>A. 46°<br><br>B. 67°<br><br>C. 115°<br><br>D. 134°​
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!