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aniked [119]
3 years ago
14

Write an equation of the line that passes through 9, 6 and is perpendicular to the line whose equation is y equals -1 over 3 x +

7
Mathematics
1 answer:
asambeis [7]3 years ago
4 0
If the line that goes through (9, 6) is perpendicular to y = -1/3x + 7, then their slopes will be opposite reciprocals.  The slope of the line given is -1/3.  The opposite reciprocal of -1/3 is +3.  If we have our new line passing through point with x coordinate 9 and y coordinate 6, we will use that x and y and the slope of 3 to solve the slope-intercept equation for b.  Like this:  9 = 3(6) + b.  9 = 18 + b and b = -9.  That means that the new equation, the one that is perpendicular to the given line, is y = 3x - 9. 
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Please help me with this <br> Thanks
alexira [117]
So you make them the same common denominator
As you see, 2/10 = 8/40
2/4 = x/40, multiply by 10
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Step-by-step explanation:

7 0
2 years ago
Prove that if {x1x2.......xk}isany
Radda [10]

Answer:

See the proof below.

Step-by-step explanation:

What we need to proof is this: "Assuming X a vector space over a scalar field C. Let X= {x1,x2,....,xn} a set of vectors in X, where n\geq 2. If the set X is linearly dependent if and only if at least one of the vectors in X can be written as a linear combination of the other vectors"

Proof

Since we have a if and only if w need to proof the statement on the two possible ways.

If X is linearly dependent, then a vector is a linear combination

We suppose the set X= (x_1, x_2,....,x_n) is linearly dependent, so then by definition we have scalars c_1,c_2,....,c_n in C such that:

c_1 x_1 +c_2 x_2 +.....+c_n x_n =0

And not all the scalars c_1,c_2,....,c_n are equal to 0.

Since at least one constant is non zero we can assume for example that c_1 \neq 0, and we have this:

c_1 v_1 = -c_2 v_2 -c_3 v_3 -.... -c_n v_n

We can divide by c1 since we assume that c_1 \neq 0 and we have this:

v_1= -\frac{c_2}{c_1} v_2 -\frac{c_3}{c_1} v_3 - .....- \frac{c_n}{c_1} v_n

And as we can see the vector v_1 can be written a a linear combination of the remaining vectors v_2,v_3,...,v_n. We select v1 but we can select any vector and we get the same result.

If a vector is a linear combination, then X is linearly dependent

We assume on this case that X is a linear combination of the remaining vectors, as on the last part we can assume that we select v_1 and we have this:

v_1 = c_2 v_2 + c_3 v_3 +...+c_n v_n

For scalars defined c_2,c_3,...,c_n in C. So then we have this:

v_1 -c_2 v_2 -c_3 v_3 - ....-c_n v_n =0

So then we can conclude that the set X is linearly dependent.

And that complet the proof for this case.

5 0
3 years ago
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