CPhill

Note if  x   = 3  and  y  = -4...they must intersect at  (x, y) =   (3, -4)  ⇒  4th quadrant

10xy  + 14x + 15y  = 166

Note that  ( 2x + 3) ( 5y + 7)   =

10xy + 15y + 14x + 21  =  166 + 21      factor the left side

(2x + 3) ( 5y + 7)  = 187

Note that  187  factors as  11 * 17

So....we can let  x = 4   and y  = 2   and we get

(2*4 + 3)  (5*2 + 7) =

  (11)     (17)  =     187

So

10 (4 * 2)  + 14(4) + 15(2)  =

80  + 56 + 30  =

80 + 86  =

166

And x + y  = 4 + 2   = 6

1/x  +  1/y  =  1/18

18 [ x + y]   =  xy

18  = xy / [ x + y ]

Let  x  = 45  and y  = 30 

18  = 45 * 30 / [ 75 ]  =   15 * 3 * 15 * 2  / [ 15 * 5]  =  90 / 5

So

x + y   =  45 + 30   =  75

In the form  by + ax + c  = 0....the slope is given by

-a / b

So.....  let -a/ b   be the slope of the first line   and since  perpendicular lines have negative reciprocal slopes, let  b /a  be the slope of the second line...so we have

-1/2   = 3 /a  

-2  = a / 3

a  = -6

See the graph, here : https://www.desmos.com/calculator/2rwoqegwpu

f (x, y)  = 2x^2  + 3y^2 + 8x - 24y +62

Take partial derivatives in x and y and set these to 0

fx   = 4x + 8  = 0    →  x  = -2

fy = 6y - 24  = 0     → y  = 4

As WolframAlpha shows....(-2,4)  is the min for this function.....and the min value  = 6

https://www.wolframalpha.com/input/?i=minimize+%5B+2x%5E2%C2%A0+%2B+3y%5E2+%2B+8x+-+24y+%2B+62%5D

Last one

We have a hexagon with a perimeter of 30m....so.....each side  is  30 / 6  =  5m

And  we  have  6  congruent triangles  each with an included angle of 60° between two equal sides of 5m

So....the total area is

6(1/2) (5)^2 sin 60°  =

3 * 25 * √3/2  =

(3/2)*25√3 m^3  =

37.5√3 m^2 ≈  65 m^2

Third one

We have decagon  .... it is composed of 10 congruent triangles each with a base of 6yd

The height of each triangle is given by  3 / tan (18°)

So...the total area  is  10 (1/2)(base) (height)  =

10 (1/2) (6) (3/ tan (18°) =

90 / tan (18°)   ≈ 277 yd^2

Second one

We have an equilateral triangle  composed of three congruent triangles...

Each  has an included angle of 120°  between two sides of 28

The  area is  3*(1/2)(side)^2* sin (120°)  =

(3/2)(28)^2 * √3/2  =  

(3/4)(784)√3  =

588√3  units^2

First one

We know the side opposite the missing angle and the hypotenuse....the trig function that relates these two is the sine

So  sine (?)  = 22 /41

And the sine inverse (arcsin) will give us the angle...so...

arcsin (22/41)  = ?  ≈ 32.45°

Since AB  is parallel to CD....then  BC is a transversal cutting two parallel lines

So.....angles B and C are supplementary [sum to 180° ]

So   

B + C  = 180

B + 3B  = 180

4B = 180

B  = 45°