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a/b =  sqrt (10)  / sqrt (21)

a  =    b sqrt (10)  /sqrt (21)

b/c =  sqrt (14)  / sqrt (15)

c/b  =  sqrt (15) / sqrt (14)

c = b sqrt (15)  / sqrt (14)

a / c   =      [ bsqrt (10) / sqrt (21)]  /  [ bsqrt (15)/sqrt (14) ] =

[sqrt (10) sqrt (14)]  /  [ sqrt (21) sqrt (15)]  =

sqrt (10) / sqrt (15)    *   sqrt (14) / sqrt (21)  =

sqrt  (10/15)  *  sqrt ( 14/21)   =

sqrt (2/3)  *  sqrt (2/3)   =

2/3

 (a/b)*(b/c) = √10/√15. If we rationalize that, we get √150/15, which is 5√6/15, which is √6/3

Very nice, textot   !!!!!!

ahhh, i see now, thanks!

The area of 4 rectangles of dimensions a*b has an area of 4ab, while a square with side length a+b has an area of (a+b)^2=a^2+2ab+b^2. We want to show:

\(4ab\le a^2+2ab+b^2\\0\le a^2-2ab+b^2\\ 0 \le (a-b)^2\)

Since \((a-b)^2\) will always be nonnegative for real numbers a and b, it will always be greater or equal to zero. \(\square\)

quarters              dimes                    nickels                      pennies

     0                        0                             0                                45

      0                        0                              1                               40

     0                        0                               2                                 35

                                               .   

                                               .       

                                               .

     0                          0                               9                          0

Can you take it from here? (Just do the same thing, except that change the values of the quarters, dimes, nickels, and pennies)

Just let   D  bisect  BC

Triangles  ADB   and  ADC  will be  under  the  same  height  and  base BD =  base DC

And triangles  under  the  same  height  and  with  the  same bases have equal areas.....

Notice  that  since AD is perp to BC ,  we  have two right  triangles  -  ADB   and ACD

The  area   of    ADB  =  (1/2)  product of  the legs =     (1/2) (AD) (BD)

Similarly.....the  area   of  ADC   = (1/2) (AD) (DC)

So.....the  combined   areas =  (1/2)  (AD)  ( BD  + DC)   =

(1/2)  (AD) (BC)  =   area of triangle ABC

I made an error! 

4/8 = √80/CQ         CQ = (8√80)/4      Wrong!!!

4/√80 = 7/CQ          CQ = (7√80) / 4      Correct!!!

5, 11, 17, 23, 29 works :)

also, quick correction, i said that the sequence has to start with a 5; it does not. It could start with a 7 (example : (7,13,19,25,31)) or 101 (example:101,9976511,19952921,29929331,39905741) ) or any prime number, but the difference has to be a multiple of 5 if the first term is not a 5.

100 from  800  to 899  inclusive

Then, additionally....

10 (9)  + 9 (9) =  171

So

100  + 171   = 

271

Since we know b = f(a), we want the "3x+6" to be equal to a.

3x + 6 = a

3x = a - 6

x = a/3 - 2

y = -3b-2

(a/3-2, -3b-2)

=^._.^=

Let triangle ABC have side lengths AB=13, AC=14, and BC=15. There are two circles located inside angle BAC which are tangent to rays AB, AC, and segment BC. Compute the distance between the centers of these two circles.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

AB = 13     AC = 14     BC = 15

s = 1/2(13+14+15) = 21

A = sqrt(21(21-13()21-14)(21-15) = 84

r = A/s = 4

Using the law of cosines, we find that ∠A = 67.380135º

∠PAY = 33.6900675º

AZ = AY = r / tan(A/2) = 6            CW = CY = 8              BW = BZ = 7

CP = sqrt(4² + 8²) = √80

4/8 = √80/CQ         CQ = (8√80)/4

PQ = sqrt(CP² + CQ²)

PQ = 18.02775638

(a) There will be a multiple of 3 if it starts with a multiple of 3.

There will be a multiple of 2 if it starts with a multiple of 2.

There will be a multiple of 5 if it starts with a multiple of 5.

There will be a multiple of 7 if it starts with a multiple of 7.

This repeats.

Thus, we only have to try (1, 3, 5), (2, 4, 6), (3, 5, 7), (5, 7, 9). The only triplet that works is (3, 5, 7).

(b) There are none. There will be a multiple of something if it gets bigger. We only try the first few, and see that they all include a number that is either 1 or composite.

Thx, MPS   !!!!

Note:

$x<-4, -4 \leq x < \frac{1}{2},\frac{1}{2} \leq x<7, x\geq 7$

$x < -4$ No Solutions

$-2 < x \leq 1$ No Solutions

$1 \leq x < 14 \qquad | \qquad x = 6$

$x \geq 7$ No Solutions

$x = 6$

No prob  !!!!

We can write

(x + 4)  +  ( 7 - x)   =   (2x  - 1)

11    =     2x  - 1

12    =  2x

x =  6

This is the only real solution as shown by the  graph here  :  https://www.desmos.com/calculator/jcjov5i1sw

The graphs intersect  at  (6,11)

I guess that works, too!

My solution was this

Since  x^2 and  y are inversely proportional, their product is constant. So,

2^2 times 10=x^2 times 4000,

x=1/10

Good work , MPS    !!!!!

Anytime!

LOL!!!!........Thx  !!!