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?
looks like it
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:
if we roll the decimal seven spaces to the right we have:
the exponent is negative.
the expression can be written as:
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:
add both of the variables:
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)
divide both sides by 7:
now substitute this into y=x+2