Here's some guidelines to help you:
For 1, 4, 5, and 6, they are 30-60-90 triangles. What's that you may ask? Well, it means that the sides have a predetermined length when one side is given. In #1 for example, you have a side of 11sqrt3, this means that "n" is 11 because in a 30-60-90 triangle, the longer side is the sqrt3 times the length of the shorter side. So to get the shorter side, we divide by sqrt3 to get 11. "m", or the hypotenuse, can be determined by taking twice the length of the shorter side. Since we figured out earlier that the shorter side is 11, 11 times 2 is 22. So the answer for #1 is n=11, m=22.
For 2 and 3, they are 45-45-90 triangles, triangles where the two lengths are the same and the hypotenuse is either leg length times sqrt2. In problem #2 for example, "y" must be 17 because one leg length is 17. "x", or the hypotenuse, is equal to 17sqrt(2) because 17 times sqrt(2).
You can apply all these rules to the other 4 problems I didn't explain.
Hope this long explanation clears your doubts!
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.)
AnswerTo the following subject of 1.08×10 in standard form you have to calculate the radius in yet to find the perimeter
SOLUTION
We have been given the equation of the decay as

So we are looking for the time
Plugging the values into the equation, we have

Taking Ln of both sides, we have

Hence the answer is 4308 to the nearest year
X = (15 - 8y)/9
-5[(15 - 8y)/9] + 12y = -107
(-75/9) + (40/9) + 12y = -107
y = -8.59
x = [15 - 8(-8.59)]/9
x = 9.3
(x,y) = (9.3, -8.59)