Linear equations substitution

Answers

Answer 1

Answer: For example, y=x+4 and y=5x+2.
That would mean that x+4=y=5x+2.
Simplify it to x+4=5x+2.
x=5x-2
-4x=-2
x=1/2.
Now, plug x into one of the equations.
y=x+4
y=1/2+4
y=9/2.
Therefore, the POI is (1/2,9/2).

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Twice a number decreased by eight is sixteen.
Sin x = 0.5What is the value of x?

WHAT IS THE NEXT VALUE 4D7G10J13

Answers

Answer:

The next letter is M and the next number is 16.

Step-by-step explanation:

Given: 4D7G10J13

It follows a pattern.

D is the 4th alphabet

G is the 7th alphabet

J is the 10th alphabet

Now we have to find 13th alphabet, which is M.

M is the 13th alphabet.

Therefore, the next letter is M and the next number is 16.

The next letter is M and the next number is 16.

The given code is 4D7G10J13

Here, the number pattern is

4+3=7

7+3=10

10+3=13

13+3=16

So, the next number is 16.

The letter pattern is

D is the 4th alphabet

G is the 7th alphabet

J is the 10th alphabet

Now we have to find 13th alphabet, which is M.

M is the 13th alphabet.

Therefore, the next letter is M and the next number is 16.

Learn more about the pattern here:

brainly.com/question/23136125.

#SPJ6

The temperature is 8 degrees F. As a cold front moves in, the temperature drops 6 degrees F per hour. What is the temperature at the end of 3 hours

Answers

Answer:

-10 Degrees

Step-by-step explanation:

Select the postulate that states a line is determined by two points.1. Postulate 1: A line contains at least two points.
2. Postulate 1a: A plane contains at least three points not all on one line.
3. Postulate 1b: Space contains at least four points not all on one plane.
4. Postulate 2: Through any two different points, exactly one line exists.
5. Postulate 3: Through any three points that are not one line, exactly one plane exists.
6. Postulate 4: If two points lie in a plane, the line containing them lies in that plane.
7. Postulate 5: If two planes intersect, then their intersection is a line.

Answers

4. Postulate 2: Through any two different points, exactly one line exists.

Postulate 2 and Postulate 1 are the only two postulates that really make sense. Postulate 2 is your answer instead of Postulate 1 for the following reason:

Postulate 1 states that a line contains at least two points.
Postulate 2 states that through any two different points, exactly one line exists.

Postulate 1 says that all lines contain at least two points, while Postulate 2 says that two points contain a line.

Answer:

Postulate 2: Through any two different points, exactly one line exists.

Step-by-step explanation:

Rebecca voluntarily decides to buy a dress that Hillary has for sale; they agree on a price of $20. Which of the following best describes who gains and who loses from the transaction?

Answers

Answer:

The question is incomplete because the options are not stated. But the right option is;

Both parties expect to gain from the transaction.

What is the distance between the points (4,3) and (0,6)

Answers

Answer:

The distance between the points (4, 3) and (0, 6) is 5 units

Step-by-step explanation:

Use the distance formula to plug in the coordinates of each point and solve

D = 5

Distance formula

Given cos alpha = 8/17, alpha in quadrant IV, and sin beta = -24/25, beta in quadrant III, find sin(alpha-beta)

Answers

Given $\cos\alpha=(8)/(17)$ , $\alpha$ is in Quadrant IV, $\sin\beta=-(24)/(25)$ , and $\beta$ is in Quadrant III, find $\sin(\alpha-\beta)$

We can use the angle subtraction formula of sine to answer this question.

$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta$

We already know that $\cos\alpha=(8)/(17)$ .

We can use the Pythagorean identity $\sin^2\theta+\cos^2\theta=1$ to find $\sin\alpha$ .

$\sin^2\alpha+((8)/(17))^2=1 \n \sin^2\alpha+(64)/(289)=1 \n \sin^2\alpha=(225)/(289) \n \n\sin\alpha=\pm(15)/(17)$

Since $\alpha$ is in Quadrant IV, and sine is represented as y value on the unit circle, we must assume the negative value $\sin\alpha=-(15)/(17)$ .

As similar process is then done with $\sin\beta=-(24)/(25)$ .

$(-(24)/(25))^2+\cos^2\beta=1 \n (576)/(625)+\cos^2\beta=1 \n \cos^2\beta=(49)/(625) \n \n\cos\beta=\pm(7)/(25)$

And since $\beta$ is in Quadrant III, and cosine in represented as x value on the unit cercle, we must assume the negative value $\cos\beta=-(7)/(25)$ .

Now we can fill in our angle subtraction formula!

$\sin(\alpha-\beta)=\sin\alpha\cos\beta-\cos\alpha\sin\beta \n\n \sin(\alpha-\beta)=(-(15)/(17)*-(7)/(25))-((8)/(17)*-(24)/(25)) \n\n\sin(\alpha-\beta)=(105)/(425)-(-(192)/(425)) \n\n \boxed{\sin(\alpha-\beta)=(297)/(425)}$

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