9514 1404 393
Answer:
(b) (1, 1/6)
Step-by-step explanation:
The "unit rate" is always the y-value for x=1 when the relationship is proportional (as it is here).
That point is not specifically shown on the graph. It would be (1, 1/6).
Answer:

Step-by-step explanation:
we would like to figure out the differential coefficient of 
remember that,
the differential coefficient of a function y is what is now called its derivative y', therefore let,

to do so distribute:

take derivative in both sides which yields:

by sum derivation rule we acquire:

Part-A: differentiating $e^{2x}$

the rule of composite function derivation is given by:

so let g(x) [2x] be u and transform it:

differentiate:

substitute back:

Part-B: differentiating ln(x)•e^2x
Product rule of differentiating is given by:

let
substitute

differentiate:

Final part:
substitute what we got:

and we're done!
40÷30s? I think that is the answer, but I'm not completely sure...
Answer:
AAS
Step-by-step explanation:
The triangles are given 2 congruent corresponding angles and 1 congruent corresponding side. So the only answer that would make sense would be the AAS Theorem.
Answer:

Step-by-step explanation:
Hello, first of all, we will check if we can factorise the polynomials.




Now, let's compute the product.

So the correct answer is the first one.
Thank you.