How do I write an equation of a line that is parallel to 2x-3y=15

1 answer:

lina2011 [118]2 years ago

4 0

The line given is -2x + 3y = 15, so we must rearrange:

=> 3y = 2x + 15
=> y = 2/3 x + 5

These two equations are EQUIVALENT, but one is in slope intercept form, the other isn't.

So in this case, gradient is 2/3 (or 2 units up, and 3 units across).

And we know that parallel lines have the same gradient. Hence, the general equation for lines parallel to it will be:

y = 2/3 x + n

where n can be any number (positive/negative, decimal/fraction/integer) except 5, which is original line.

For example, y = 2/3 x + 2

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Can someone help with 2B, please

Hunter-Best [27]

Answer:

f'(1)=150ln(1.5)

Step-by-step explanation:

I'm not sure why you would need a table since the limit definition of a derivative (from what I'm remembering) gives you the exact formula anyway... so hopefully this at least helps point you in the right direction.

My work is in the attachment but I do want to address the elephant on the blackboard real quick.

You'll see that I got to the point where I isolated the h's and just stated the limit equaled the natural log of something out of nowhere. This is because, as far as I know, the way to show that is true is through the use of limits going to infinity. And I'm assuming that you haven't even begun to talk about infinite limits yet, so I'm gonna ask you to just trust that that is true. (Also the proof is a little long and could be a question on it's own tbh. There are actually other methods to take this derivative but they involve knowing other derivatives and that kinda spoils a question of this caliber.)