Question

medal will be rewarded for whoever ha correct Algebra 1 answer

Answer 1

Is it clear enough that you should assume n = 0, or n = 1 for the start of 3?

Answer 2

No, because 3^0 = 1. So 3^(0)*5 = 5. Your initial digit would definitely be five.

Answer 3

Yes. (Sorry, didn't intend to confuse you.) As for number four: |-1|^2 is still just one. So f(2) = 2*|-1|^2 = 2*(1)^2 = 2*1 = 2.

Answer 4

And as a general rule, one put to any power will always be one. Your first answer is correct; it's a geometric sequence. In your second one, even if n = 1, none of the answers properly align to that; This would be the set resulting from either n = 0 or n = 1:\[n _{0} = 5*3^{0} = 5*1 = 5\]\[n _{1} = 5*3^{1} = 5*3 = 15\]\[n _{2} = 5*3^{2} = 5*9 = 45\]\[n _{3} = 5*3^{3} = 5*27 = 135\]|||-|||-|||-|||

Answer 5

Yes, two and three are correct, and four should be \[(2)*|-1|^{2} = 2*(1)^{2} = 2*1 = 2\]

Answer 6

thank you again for explaining! Can you help me one one more please?

Answer 7

No problem, and sure.

Answer 8

*one more

Answer 9

Okay, so since you have a set of points, you can do a few things to test this: You ca find the slope of the line that may or may not be produced, and test whether the slope remains the same. I'm sure you know the slope formula. \[\frac{ y _{2}-y _{1} }{ x _{2}-x _{1} }\]From here, just plug in the coordinates \[(-2, -7)|||(-1, 1)|||(0, 8)|||(1, 17)|||(2, 15)\]...\[\frac{ (-7)-(1) }{ (-2) - (-1) } = \frac{ -8 }{ -1 } = 8\]
Now test if this is the same as slopes between other points:
...\[\frac{ (1)-(8) }{ (-1)-(0) }= \frac{ -7 }{ -1 } = 7\]
A linear function should have a slope that is consistent regardless of where you sample it. Does this function have the same apparent slope?

Answer 10

So far, yes

Answer 11

Are you sure? Look at the results I got. Are those the same?

Answer 12

Exactly!

Answer 13

thank you!!!!

Answer 14

Glad I could help. If you have any other questions, just ask. I'm sure that plenty of people will be able to help.