CPhill

x^2 + 25x    + c

Take  (1/2)  of 25   =  25/2

Square this  =    625/4   =   c

\(\frac{-5\pm i\sqrt{11}}{6}\)

3x^2 + 5x + k

The product  of the two  roots  is

(-5 + i√11) / 6  *    (-5 - i√11) / 6    =    (25  - i^2* 11)/ 36 =   (25 - (-1)(11) ) / 36  =

[25 + 11] / 36    =   36 / 36   =    1

And this product will  equal  k/3

So

k/3  =  1         multiply both sides by 3

k  = 3

-4 < 2(x - 1) < 8          divide through by  2

-2 <  x  - 1 <  4           add 1 to everything

-1 < x  < 5

So

a + b  =

-1  +  5  =

4

Since  x , y  ≥ 0.......we are only concerned with the lattice points intersected by the line in the first quadrant

So  we  have

y  = -2x + 18

The line has a y intercept  of  18.....so one lattice point  is  (0,18)

And letting  y  = 0, we can find the x intercept as

0 = -2x + 18

-18  = -2x

9  =  x ....so  a second lattice point is (9,0)

Now.....the slope  is  - 2.....which  means that y will change by -2  when x changes by 1

So.....the other lattice points intercepted by the line are

(1, 16)    (2, 14)   (3, 12)  (4, 10)  (5, 8)  (6, 6)  ( 7, 4)  (8,2)

So....10 lattice points

See the graph here :

https://www.desmos.com/calculator/climjgy2hb

7x^2  + 5x  = h

7x^2 + 5x - h  = 0

We will have one solution when the discriminant  =  0

So

5^2  - 4(7)(-h)   = 0

25 + 28h = 0

25 =  - 28h

-25/28 =  h

This graph shows that this is true :

https://www.desmos.com/calculator/jbpjhjdgt4

y^2  +  24y + 16 + k

Take  (1/2) of 24  =  12

Square this  = 144

So....this must be added to y^2 + 24y   to have a perfect square binomial

Which implies that

144 = 16 + k          subtract 16 from both sides

128  =  k

tan(46°)  = x / 12     multiply both sides by 12

12* tan (46°)  = x ≈  12.4

Let the height of the pyramid  =  h

And let the distance from the face of the base to the point directly beneath the apex = x

So....we have two equations

tan(36.4)  =  h / (350 + x)       multiply both sides by (350 + x)

(350 + x) tan (36.4)  = h      (1)

And the second one is

tan (51.9)  = h / x

x* tan(51.9)  = h     (2)

Equating (1)  and (2), we have

(350 + x) * tan(36.4)   =  x * tan (51.9)     simplify

350*tan(36.4)  + x * tan (36.4)  = x * tan (51.9)

350* tan(36.4)  =    x* tan(51.9) - x * tan (36.4)

350*tan (36.4)  = x [ tan(51.9) - tan(36.4) ]         

350 tan(36.4) / [ tan (51.9) - tan(36.4 ) ]   = x

x ≈ 479.558 ft

Using (2)  we can find the height as

479.558 * tan(51.9)  =  h ≈  611.6 ft

The distance, d, that the submarine descends in one minute can be expressed as

sin(5°)  =  d / 792      multiply  both sides by  792

792*sin(5°)  =  d

So...to find out how many minutes it will take the sub to reach a depth of 200 ft we can evaluate this

200 / [ 792*sin (5°) ]  ≈ 2.9 min ≈ 2 min 54 sec 

Using the Law of Sines

x / sin (32°)  = 11/ sin(47°)

Multiply both sides by  sin(32°)

x  =  11 * sin(32°) / sin(47°)