CPhill

(x - 2) / 3  +  (x + 5) / 2   =  1/3          multiply through by the common denominator of 2, 3  = 6

2 (x -2)  + 3( x + 5)  =  2          simplify

2x  - 4  + 3x + 15  = 2             combine like terms

5x + 11  = 2                          subtract 11 from  both sides

5x  =  - 9                            divide both sides by 5

x = -9 / 5

Let's look at this a little closer......I don't know if "Point A"  is some specific point   or any random point on the circle.....but....if it's a random point.......look at this graph

Let  " Point  A "  be on the on the circle   =  ( a , sqrt (r^2 - a^2))

The slope of  the tangent line to this circle will be   =    -a / sqrt (r^2 - a^2)

So....the equation of a  line tangent to  the circle at  this point  is

y =  [ -a / sqrt (r^2 - a^2)] ( x - a) +  sqrt (r^2  - a^2)

To find "Point P"  let  y  = 0   we have that

0 =   [ -a / sqrt (r^2 - a^2)] ( x - a)  +  sqrt (r^2  - a^2)

 [ a / sqrt (r^2 - a^2)] ( x - a)   =   sqrt (r^2  - a^2)

a (x - a)   =   r^2 - a^2

ax - a^2  =  r^2  - a^2

ax =  r^2

x  = r^2 / a   ...   so  "Point P"   (  r^2/ a , 0 )

So.......  the   the base of triangle OAP will  be  r^2 / a      and the height will be sqrt ( r^2 - a^2)

So....the area of this triangle will be    (1/2) (r^2 / a) sqrt (r^2 - a^2)  =    (r^2) sqrt (r^2 - a^2) / [ 2a ]

So....suppose  r  = sqrt (50)  and  a  = 4

Then the area of triangle OAP  will  be    50 sqrt (34) / 8  ≈  36.44 sq units

But suppose that   r =sqrt (50) and  a  = 2

Then the area of triangle OAP  will be  50 sqrt ( 46) / 4  ≈ 84.78 sq units

It might be interesting to see if we can discover a pattern here.

Note

Three elements....two at  time  (1,2) (1,3) (2,3)....sum  =   12

Four elements .... two at a time  (1,2) (1,3) (1,4) (2,3) (2,4) (3,4)  .....sum  =  30

Five elements....two at a time  (1,2) (1,3) (1,4) (1,5) (2,3) (2,4) (2,5) (3,4) (3,5) (4,5)  ....sum  = 60

Note that for N elements taken  2 at a time......the pattern of sums seems to be

(N - 1) ( N ) ( N+ 1)  / 2  

So....for 6 elements taken 2 at a time.....the sum should be

(5) (6) (7)  / 2  =     210 / 2  =   105

Verify for yourself that this is true.......!!!!!!

Since AO is ll to QR then PR is a transversal that cuts parallel lines

Thus angle POA  = angle PRQ

And angle QPR  = angle APO

Thus, by Angle-Angle congruency....triangle QPR  is similar to triangle APO

But similar triangles are similar in all respects.....

So........  PQ / PR  = PA / PO    which also implies that     PA / PQ  = PO / PR

But.....PA  = AQ.....so.....PA  = (1/2)PQ

Then....by substitution.......

(1/2) PQ / PQ  =  PO / PR

Simplifying.....we have that    

(1/2)  = PO  / PR    whicih implies that

PO  = (1/2)PR

And since PR is the diameter of the circle and PO is (1/2)  of PR.....O must be the center of the circle

28 + 30 + 32 + .... + 86

Note....that we can pair the terms as follows

[ 28 + 86 ] + [ 30 + 84] + [ 32 + 82 ] +......+ [56 + 58]

Each pair sums to 114

And the number of pairs is given by   [56 - 28] / 2 + 1  = 15 pairs

So...the sum is   15 * [ 114 ]  =    1710

 (6/13)𝑥 = 51     multiply both sides by  13/6

x  = 51 (13/6)   =  663 / 6

45.1 + 21.5𝑥 =  125.6      multiplly through by 10 to clear the decimals

451  + 215x  = 1256       subtract 451 from both sides

805  = 215x                  divide both sides by 215

805 / 215  = x  = 161 / 43

Matt wants to walk his dog around the block. The block has a rectangle shape. The dimensions of the block is 3 miles long and 2450 yards wide. How many feet would Matt walk if he walked his dog around the block two times? Write an equation and solve for x.

There really isn't any "x"  to solve for....we  just need to convert miles and yards to feet

1 mile  = 5280 ft....so   3 miles   = 3 * 5280 =  15840 ft

1 yard  = 3 ft.......so 2450 yards  = 2450 * 3  = 7350 ft

One complete trip comprises the perimeter of a rectangle that is   15840 x 7350

So......one trip around the block is equal to the perimeter of this rectangle  = 2 [ 15840 + 7350 ]  = 46380 ft

So....two trips is twice as far   = 2 * 46380  = 92760 ft

A number to the second power is 7 less, than the quotient of forty-two over 8. Write an equation to represent the problem.

Let N be the number.....so we have that

N^2  +   7  =  42 / 8

134x+(-100)= 112, Solve for x

Add 100 to both sides

134x  = 212      divide both sides by 134

x  = 212 / 134  =    106 / 67

 5 - (2x + 3)  =

5 - 2x  - 3   =

2 - 2x

-2x^2 - 5x = -50       multiply through by -1

2x^2  + 5x  = 50      complete the square on x

2 [x ^2  +  (5/2)x ]  = 50       divide both sides by 2

x^2  +  (5/2)x     = 25         take (1/2)  of (5/2)  = 5/4.....square this = 25/16,  and add to both sides 

x^2  + (5/2)x  + 25/16  =   25  +  25/16         factor the left side and simplify the right

[ x +  5/4 ] ^2   =   [ 400 + 25]  / 16          take both roots

x + 5/4  =  ±√ [ 425]  / 4

x + 5/4  =   ±5√17 / 4       subtract 5/4 from both sides

x  = ±5√17 / 4  - 5/4  =  [  -5  ± 5√17 ] / 4  

 0.3939   this is a rational number because it can be expressed as the ratio  3939 / 10000

√ 45  =   3√5        any  number that has a square root remaining after we simplify it is irrational

9      is  rational [ can be expressed as a ratio such as 18/2, etc.]......also an integer    ....also a whole number [ a "counting number" or  0]

-2  is rational  [ can be expressed as the ratio such as   -4/2 , etc] ......also an integer ...it is not a whole number because it is not a "counting" number or 0

Perimeter  =  2 (W + H)

So....we have this

2 ( W + H)  < 180       sub in for the width

2 ( 26  + H)  < 180        divide both sides by 2

26 +  H  <   90               subtract 26 from both sides

H <  64 in

So....for the perimeter to be less that 180 in, the height must be less than  64 in