Answer:
Given statement is TRUE.
Step-by-step explanation:
Given that line segment JK and LM are parallel. From picture we see that LK is transversal line.
We know that corresponding angles formed by transversal line are congruent.
Hence ∠JKL = ∠ MLK ...(i)
Now consider triangles JKL and MLK
JK = LM {Given}
∠JKL = ∠ MLK { Using (i) }
KL = KL {common sides}
Hence by SAS property of congruency of triangles, ΔJKL and ΔMLK are congruent.
Hence given statement is TRUE.
The answer is '<span>f(x) is an odd degree polynomial with a positive leading coefficient'.
An odd degree polynomial with a positive leading coefficient will have the graph go towards negative infinity as x goes towards negative infinity, and go towards infinity as x goes towards infinity.
An even degree polynomial with a negative leading coefficient will have the graph go towards infinity as x goes toward negative infinity, and go towards negative infinity as x goes toward infinity.
g(x) would have a a positive leading coefficient with an even degree, as the graph goes towards infinity as x goes towards either negative or positive infinity.
</span>
has gradient
which at the point (-1, 4, 3) has a value of
I'm not sure what the given direction vector is supposed to be, but my best guess is that it's intended to say 
, in which case we have
Then the derivative of at (-1, 4, 3) in the direction of 
is
Answer:
Option D is the right option.
Explanation:
Observe from the graph that the value of X starts at X=0 and the graph is going to the right infinitely.
So the domain of the function should be:
Hope this helps...
Good luck on your assignment..
