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saw5 [17]
3 years ago
10

For what values of k the relation R:{(k2 , 5), (3k, 6)} is not a function?

Mathematics
1 answer:
Nastasia [14]3 years ago
7 0

Answer:

k=0 and k=3.

Step-by-step explanation:

To this relation be a function, the horizontal coordinates can't be equal. So, let's find the value of k that makes those coordinates equal.

k^{2}=3k\\ k^{2}-3k=0\\k(k-3)=0\\

Using the null factor property, we have

k=0\\k-3=0 \implies k=3

Therefore, the given relation is not a function when k=0 and k=3.

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What is the nth term for the sequence 5 , 14 , 29 , 50 , 77
Marina86 [1]
The answer would be 3n^2 + 2.

This can be found/proven by replacing "n" with term number (1,2,3,4...), then solving to get the final number. For example 3 * 1^2 + 2. You would first do 1^2, which is 1. Next, you would multiply 1 by 3, to get 3. Finally, you'd and the 2 to get 5. 5 is the 1st term, and you can use this same equation to get the rest of the terms you need.

I hope this helps!
8 0
3 years ago
Find the circumstances of the circle. Use pi = 3.14
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Step-by-step explanation:

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5 0
3 years ago
A linear function is parallel to the line y= 4x + 8 and also goes through the point (2,15).What is the linear equation for this
horrorfan [7]

When written in the y=mx+q form, the slope of a line is given by the coefficient m. Moreover, two lines are parallel if the have the same slope.


Now, the slope of the known line is 4, so our line's slope will be four as well.


In general, when you know the slope m of a line and one of its points (x_0,y_0), the equation of the line can be derived from the following formula:


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