CPhill

When cosx = 0 , sin(cos(x)) will = 0

And cosx = 0  at  pi/2 and 3pi/2   (restricting ourselves to [0,2pi])

2x^2-x+5,g(x)=x+1    assuming that the first function is f(x)

[f+g](x) means that we just add the functions together  = 2x^2 + 6

[f + g](6)  means that we "plug" 6 into what we've just found in the first part......so we have...... 2(6)^2 + 6  = 2(36) + 6 = 72 + 6 = 78

That IS excellent....very clever !!!!!!

3. P=(1.2w)/h^2, find h.

Multiply both sides by h^2

h^2*P = 1.2w

Divide both sides by P

h^2 = (1.2w)/P

Take the positive root of both sides...since h is a positive quantity

h =√[1.2w/P]

4. 9/10g=7+2/3k, find k

Multiply both sides by the LCM of 3 and 10, i.e., 30

27g = 210 + 20k

Subtract 210 from both sides

27g - 210 = 20k

Divide both sides by 20

[27g - 210 ]/ 20 = g

5. 2-3/4p=5/6r+5, find r.

Multiply through by the LCM of 4 and 6, i.e., 12

24 - 9p = 10r + 60   

Subtract 60 from both sides

24 - 9p - 60 = 10r

Divide both sides by 10

[24 - 9p - 60 ] / 10  = r

Notice that the trick of multiplying through by the LCM first allows us to get rid of those nasty fractions in our work !!!

150% = 1.5     "of" means multiply ......so we have

1.5 x 74 =  111

Notice that 

40.8  =   40  + .8  ......and dividing each thing by 8  =   5 + .1    =  5.1

That settles it....... 117 is the Most Common Multiple (answer)........

3 = 3 x 1

39 =13 x 3

117 = 3² x 13

So taking each different factor with the higest power, we have

3² x 13  = 117   = LCM

OK...I misread that....so we have....

A =  [a (v_f  - v_i ) ] / t       multiply both sides by t

At = [a (v_f  - v_i ) ]           divide by a

(At) / a =  v_f  - v_i             add v_i to each side

[(At) / a ] + v_i =  v_f

We could also use the Law of Cosines to find this. Note that the supplementary angle to "15° northward" is 165°

So we have

√[80² + 60² - 2(80)(60)cos(165°)] = 138.83 miles