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OK, parpar1.....

We  are looking for the  minimum (squared) distance  from the origin to the line   3x + y  =15

Let this point  be   (x  , 15 - 3x )

D^2  =  ( x- 0)^2  +  ( 15  -3x-0) ^2

D^2  =  x^2  + ( 15-3x)^2

D^2  = x^2  + 225 - 90x + 9x^2      take the derivative and set to 0

D'^2  =  2x  -90 + 18x   = 0

20x - 90  = 0

20x  =90

x= 9/2

y = 15 - 3(9/2)  = 30/2 - 27/2  =  3/2

So x^2  + y^2   is a minimum  when  x= 9/2  and y =3/2

The minimum  is   (9/2)^2  + (3/2)^2   =   [ 81 + 9 ] / 4  =   90 / 4 =  45/2  =  22.5

Thanks so much, Cphill!

1

(a)  the populauion  consists  of  the guests staying at the hotel

(b)  the poulation  consists of the  teachers  at the school

(c)  the population consists of  the customers of the business

a = 7   b  =  5    c  = 5

The circumradius length is  given by:

     abc

_________________________________________  =

√[ (a + b + c) ( a + b - c) (a + c - b ) ( b + c - a)]

7* 5 * 5

________________________________________ =

√ [ ( 7 + 5 + 5)  ( 7) ( 7)  ( 3) ] 

     175

_____________   =

√[ (17) (49) (3) ]  

     175

_______________  =

  7 √51

25 / √51

Using the Law of Cosines, we can  calculate  angle BAC

7^2  - 5^2  - 5^2

_____________   = cos BAC

     -2(5)(5)

1 / 50  =cos BAC

arccos (1/50) =  BAC  = 88.85°

And  since BAC  is an inscribed angle, then central angle BOC  is twice this  = 177.7°

So the area of   BOC =  (1/2) ( 25/ √71)^2 sin (177.7)  ≈   .177  units^2 

Look at the  graph ,here :  https://www.desmos.com/calculator/grfdjntjey

It appears that just  G and I  satisfy the inequalities

IS it just me or can I not see the picture? I think theirs a picture, but I cannot see it. Can anybody else?

slope of p  =  [-5 -2]  / [ -3 - -1] =  -7 / -2  = 7/2

slope of a perpendicular line has a negative reciprocal slope to p  =    -2/7

Area of equilateral triangle = 

(1/2) ( side)^2 sqrt (3) / 2   = 64 sqrt (3)     multiply both sides by 4 and divide by dqrt (3)

side^2  = 256

16 cm   = side

Area  when side is decreased by 2 cm

(1/2) ( 14)^2 sqrt (3)/2 =

196 sqrt ( 3)/ 4  =

49 sqrt (3)

Area is decreased  by   (64 - 49) sqrt (3)   =  15 sqrt 3)  cm^2

Angle ACB = 180 - 142

Let the length of diagonal be 10

Square area is 10²/2 = 50

The diagonal and 2 sides of a rectangle form a 30-60-90 triangle.

The lengths of  both legs are         sin30º * 10 = 5             sin60º * 10 = 8.660254038

The area of the rectangle is             5 * 8.660254038 = 43.30127019

R² = (43.30127019 / 50)²     ==>  R² = 0.75    or   3/4

There are 46 different products that can result.

In this case, the center of dilation is, at the same time, the centroid of both triangles.

The triangle DEF is turned upside-down, and that makes the scale factor negative. 

The scale factor is -1/2

Dilations preserve lengths. Therefore XY=12.

The answer is 22.

2.  The only points are A and I.

Y'Z' = 37.10795063

1: AIGE

2: idk I'm only in 8th grade.

Let's find what value(s) of  x  make  f(x) = 8

f(x)  =  8

x² - 6x + 4  =  8

x² - 6x - 4  =  0

By the Quadratic formula,

\(x\ =\ \frac{6\pm\sqrt{(-6)^2-4(1)(-4)}}{2(1)}\ =\ \frac{6\pm\sqrt{52}}{2}\ =\ \frac{6\pm2\sqrt{13}}{2}\ =\ 3\pm\sqrt{13}\)

And so....

\(f(3+\sqrt{13})=8\)       and       \(f(3-\sqrt{13})=8\)

First let's plug in  3 + √13  for  x  into the function  g( f(x) )  =  2x + 3

\(g(\ 8\ )\ =\ g(\ f(3+\sqrt{13})\ )\ =\ 2(3+\sqrt{13})+3\ =\ 9+2\sqrt{13}\)

Next let's plug in  3 - √13

\(g(\ 8\ )\ =\ g(\ f(3-\sqrt{13})\ )\ =\ 2(3-\sqrt{13})+3\ =\ 9-2\sqrt{13}\)

And so the sum of all the possible values of  g(8) is:

\((9+2\sqrt{13})\ +\ (9-2\sqrt{13})\ =\ 18\) .

y² + 28y + 16 + k   will be the square of a binomial when:

16 + k  =  (1/2 * 28)²

16 + k  =  (14)²

16 + k  =  196

k  =  180

Check:

y² + 28y + 16 + 180   =   y² + 28y + 196   =   (y - 14)(y - 14)   =   (y - 14)²

7x²  + 8x  =  h       Let's subtract  h  from both sides to get

7x² + 8x - h  =  0

This will have only one root/solution when the discriminate is 0,  that is, when:

8² - 4(7)(-h)  =  0

64  +  28h  =  0

Can you finish solving the above equation for  h  ?

Whatever value of  h  we get from that equation is the value of  h  which gives us exactly one solution for  x .

b)  qualitative

c)  discrete

d)  continuous

First let's look at  △ABC.  By the Pythagorean Theorem,

AC²   =   AB² + BC²   =   ( 3√2 )² + ( 3√2 )²   =   18 + 18   =   36    and so    AC  =  6

△ABC  is a right isosceles triangle which means the base angles are both  45°

And so   m∠CAD  =  m∠BAD - m∠BAC   =   75° - 45°   =   30°

Now we can see that  △ACD  is a  30-60-90 triangle,

and in all 30-60-90 triangles, the side across from the 30° angle is half the hypotenuse.

In this case, that means   CD  =  1/2 * AC   =   1/2 * 6   =   3