CPhill

(4! + 4/4) * 4 =

(24 + 1) * 4  =

(25) * 4  =

100   !!!

This is a 30° in the first quadrant ....and since both x and y are positive, it's also a 30° angle in the 3rd quadrant (i.e., 210°)......of course all co-terminal angles (±n2pi) to these will have the value......

To find the period......look at the distance from one of the highest "peaks" to the next one......note that one highest "peak" occurs just before 0 and the next one occurs just before "pi."  So the period appears to be "pi.'  [Note that "peak" isn't strictly a "math" term, but it seems to work OK, here !!!]

It's Graph "C."   The function can be written as ....tan 1/2(x).... where the "1/2" makes the normal period of the tangent graph twice as long......!!!!

This is a graph of the normal sine function "flipped over" with an amplitude of 2 and shifted to the right by pi/2 units. So, since the normal sine graph has x intercepts at 0 ± n2pi and pi ± n2pi, this graph will have x intercepts at pi/2 ± n2pi and [(3 pi)/ 2] ± n2pi ....where n is just an integer

Yes, 70.52° is correct.....!!!

1 + x = √(1 + 5x)   square both sides

1 + 2x + x ^2   = 1 + 5x     subtract 1 + 5x from both sides

-3x + x^2  = 0     rearrange

x^2 - 3x = 0        factor

x (x - 3)     and setting each factor to 0, we get that  x = 3  or x = 0

Make sure to check these answers....square root equations of this sort often give "extraneous" answers  that will not work in the original problem......here, we're OK though !!!

Either we have

[1-(-49) / 81 =

50/81

Or we have

1 - (-49/81) =

1 + (49/81) =

[81 + 49] / 81 =

130/ 81

c^2= 8.9876 x 10^16

So we would have, in standard form.....

c^2 = 89,876,000,000,000,000  ..... and we could solve this by taking the square root of both sides

c =  about  ± 299793262.1

Thanks for those laughs, rosala......the "speed limit" sign was great....also " the hard homework" with the dog  !!!!!