CPhill

What equations???........there are none shown....

We can use the Law of Sines here

c/sin(67) = 12/sin(34)        ......multiply both sides by sin(67)

c = 12*sin(67)/sin(34) = about 19.8

3² + 4 + 5 = 

9 + 9 =

18

Another possibility

4² + (5 - 3)

16 + 2

18

[There are probably others, I'm sure !!!! ]

Let's trace this one through !!!

We know 2 side lengths

A is not greater than 90

Calculate h = bsinA = 12sin(35) = 6.88

So a > h   and a < b

So the answer is .....2 triangles, solve both

We can use the Law of Cosines

17^2 = 11^2 + 14^2 - 2(11)(14)cos(L)

289 = 121 + 196 - 308cos(L)

-28 = -308cos(L)        divide both sides by -308

(28/308) = cos(L)        take the cosine invese

cos^-1 (28/308) = about 84.8°

We can use the Law of Cosines here

a^2 = 7^2 + 9^2 - 2(7)(9)cos(46°)

a^2 = 49 + 81 - 126(.695)

a^2 = 42.43        ......take the square root of both sides

a = about 6.5

g(t)=t(4-t)^1/2        we can use the product and chain rules here

g'(t)  = (4-t)^1/2  + (1/2)t(4-t)^(-1/2)(-1) = (4-t)^1/2 - (1/2)t(4-t)^(-1/2) =

.......(factoring).....   (4-t)^(-1/2) [ (4-t) - (1/2)t]   =  (4-t)^(-1/2)[4- (3/2)t ] =

(4-t)^(-1/2) [(8 -3t)/2]

Here is one of the best pages on the internet that covers Phi (and it's reciprocal, phi):

∫(1/x) dx =  ln x  + C

Note that if we differentiate ln u + C, we have  .......(1/u)du....and if we substitute x for u, we have (1/x)(1) = 1/x....which is the fuction in the original integrand.... !!!!