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The slope of this line is \(-5\). In order for two lines to be perpendicular, their slope's must have a product of \(-1\). So \(-5x = -1\), solving this we can get \(x = \frac{1}{5}\). \(\frac{1}{5}\) is your slope.

- Daisy 

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

The question states that the price, \(x\) is not less than \(100\), which means it has to be greater than \(99\). But, it also did not say that it had to be greater than \(100\), so \(x ≥ 100\).

- Daisy

Post only one question at a time and learn from the answers.

WE are not here to do all your homework for you !

Plus none of these are in fact questions! You have not said what you even want !! 

A vital part of the question is missing !!

Thank you for a nice explanation, CPhill!

Yeah. I also probably should have explained my answer like you did. It's like the saying "give a man a fish and he will feed himself for a day, teach a man to fish and he will feed himself for a lifetime". Same thing here

I got that also, hopefully we are both correct rather than both incorrect... Haha.

- Daisy

I got f(-2)= -38 but I might have gotten it wrong so ask another person for the answer. Its always best to have more than one perspective anyway

Since in this case \(x = -2\), then we can subsitute \(-2\) in for all the \(x\)​'s. We can get \(4{(-2)}^{3} - 4{(-2)}^{2} + 10\), which equals \(-32 - 16 + 10\), which equals \(-38\).

- Daisy

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

Thanks! Your correct!

No, I don't want to do that! I want to really understand the question!

Yes, correct! Thanks!

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

Hint: Find the slopes of both graphs, which in this case: -1/2, 2. 

Thanks! Very good!

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

I never suggested mine was the full answer. It was an rough outline of the answer that I had in front of me. 

Since it was never intended as a full answer, and I did make that quite clear, I certainly do not need to be excused.

I was well away that M was in place before it was discovered to be the midpoint and I would have explained that if I had been asked.

4 sided quadrilaterals with parallel sides can also be irregular, and they usually are.

The only 4 sided quadrilateral that is regular is a square.  Since it is the only one with all sides and all internal angles equal.

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

Thank you, CPhill!

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°