The derivation in example 6.6.1 shows the taylor series for arctan(x) is valid for all x ∈ (−1,1). notice, however, that the series also converges when x = 1. assuming that arctan(x) is continuous, explain why the value of the series at x = 1 must necessarily be arctan(1). what interesting identity do we get in this case?

answer:

looks like it

step-by-step explanation:

a reflection then translation.

where is the triangle?

for this case we must find a reasonable estimate, in scientific notation, for a wavelength given by:

0.000000492 meters

if we roll the decimal seven spaces to the right we have:

the exponent is negative.

the expression can be written as:

answer:

option c

answer:

(-6, -4)

step-by-step explanation:

use substitution.

since y is already isolated, substitute it into the y of: 4x+3y=-36.

you should get: 4x+3(x+2)=-36

simplify this and you'll get:

4x+3x+6=-36

add both of the variables:

7x+6=-36

add -6 to both sides, (because there is a -36 on one side, and a 6 on the other, you have to add a negative to eliminate the 6 whilst adding to the -36)

7x=-42

divide both sides by 7:

x=-6

now substitute this into y=x+2

y=-6+2

y=-4