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Yuki888 [10]
2 years ago
10

WILL MARK BRAINIEST

Mathematics
1 answer:
antiseptic1488 [7]2 years ago
3 0
A) (0,1), (1,-2), (-1,4), (2,-5)
b) (0,4), (-1,2), (-2,0), (1,6)
c) (-2,-7), (-1, -7), (0,-7), (1, -7)
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Letters A, B, D and E represent numbers on the number line. The number line has evenly spaced marks
Alex787 [66]

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0.5

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Question 18 of 25
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2 years ago
Write in slope-intercept form the equation of the line passing through the given point and perpendicular to the given line.
Ksivusya [100]

Answer:

y = x + 1

Step-by-step explanation:

First you find the slope of the first line using the rise over run method. Here you can see that the line goes up 1 and to the right one in order to reach the next point making the slope 1x or just x. Then you need to make it so that the line intercepts the point by changing the y intercept to 1 making the equation y = x + 1.

I hope this helped!!

4 0
2 years ago
Will mark BRAINLIEST if correct! HURRY!!!!
laiz [17]

Answer:

all real numbers greater than or equal to -1

Step-by-step explanation:

3 0
3 years ago
Determine whether the set of all linear combinations of the following set of vector in R^3 is a line or a plane or all of R^3.a.
Temka [501]

Answer:

a. Line

b. Plane

c. All of R^3

Step-by-step explanation:

In order to answer this question, we need to study the linear independence between the vectors :

1 - A set of three linearly independent vectors in R^3 generates R^3.

2 - A set of two linearly independent vectors in R^3 generates a plane.

3 - A set of one vector in R^3 generates a line.

The next step to answer this question is to analyze the independence between the vectors of each set. We can do this by putting the vectors into the row of a R^(3x3) matrix. Then, by working out with the matrix we will find how many linearly independent vectors the set has :

a. Let's put the vectors into the rows of a matrix :

\left[\begin{array}{ccc}-2&5&-3\\6&-15&9\\-10&25&-15\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix  ⇒

\left[\begin{array}{ccc}-2&5&-3\\0&0&0\\0&0&0\end{array}\right]

We find that the second vector is a linear combination from the first and the third one (in fact, the second vector is the first vector multiply by -3).

We also find that the third vector is a linear combination from the first and the second one (in fact, the third vector is the first vector multiply by 5).

At the end, we only have one vector in R^3 ⇒ The set of all linear combinations of the set a. is a line in R^3.

b. Again, let's put the vectors into the rows of a matrix :

\left[\begin{array}{ccc}1&2&0\\1&1&1\\4&5&3\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

\left[\begin{array}{ccc}1&1&1\\0&1&-1\\0&0&0\end{array}\right]

We find that there are only two linearly independent vectors in the set so the set of all linear combinations of the set b. is a plane (in fact, the third vector is equivalent to the first vector plus three times the second vector).

c. Finally :

\left[\begin{array}{ccc}0&0&3\\0&1&2\\1&1&0\end{array}\right] ⇒ Applying matrix operations we find that the matrix is equivalent to this another matrix ⇒

\left[\begin{array}{ccc}1&1&0\\0&1&2\\0&0&3\end{array}\right]

The set is linearly independent so the set of all linear combination of the set c. is all of R^3.

4 0
2 years ago
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