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OOPS...just a sign error....now our answers agree

Thanks for the  correction, jugoslav !!!

If f(0)   = 7.....then  d   = 7

And  we  have this system

a  +   b   +   c    +   7      =  10     ⇒     a  +   b   +   c     =  3

8a + 4b +  2c      + 7     =   15     ⇒    8a  + 4b  +  2c   =  8

27a + 9b + 3c  +   7    =   28    ⇒       27a  + 9b + 3c   =  21

Multiplying the  first equation by   -2   and adding to the second  gives  us

6a + 2b  =  2       (4)

Multiplying  the first equation by - 3  and adding to  the  third produces

24a   + 6b    =  12    (5)

Mutiply ( 4)  by -3  and add to (5)  and we  get that

6a  =  6

a  =  1

6(1) + 2b   = 2

2b = -4

b  = -2

1   - 2   +  c    =  3

-1  + c  =  3

c   = 4

a + 2b + 3c + 4d   =   

1 - 4 + 12  +  28   =

37

Say that the length of the radius of the large quarter circle is 4. That would mean we can solve for the red portion. If break it down, we can see that it is simply a semicircle with a right isosceles triangle inside with the legs having a length of 2. The area of the triangle would be 2. if you add half of the red part, you get a quarter-circle with radius 2, which would have an area of pi. Subtracting one from the other we get that the area of half of the red part if pi-2. So the whole thing is 2pi-4 The two yellow semi-circles are just 1 circle with the raidus being half of 4, or 2. that would be 4 pi. But a quarter circle of radius 4 would have an area of 4pi. The red part accounts for this as it is repeated, so we subtract that from the semicircle. But the difference between the large quarter circle and the semi-circles would be the amount you subtracted, 2pi-4. So the red and blue areas are the same.

PA must be shorter than PT!!!

A, B, and T are three points lying on a circle.  If PT is a tangent line of the circle and AB = 7 and PT = 12, then what is PA?

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

PT² = PA * PB         PA => x

144 = x (x + 7)

x² + 7x - 144 = 0

x = 9

We can use  the tangent-secant theorem to solve this

PT^2   = AP ( AP + AB)

( 12)^2  =  AP^2  +  AP7

144  = AP^2  +  7AP      rearrange as

AP^2  + 7AP    -  144    =     0          factor as

(AP  +16)  (AP - 9)   =  0

The first factor set to  0   will give us what we need

AP  - 9    =    0

AP    =   PA   =  9

6x = 35/x  - 42x^2        multiply through  by   x^2

6x^3  = 35x  - 42         rearrange  as

6x^3  - 35x  +  42    =  0     

This  only  has one real  solution    x  ≈ -2.8754

The  complex solutions are

x ≈  1.4377  - .6062 i      and      x ≈   1.4377 + .6062 i

Very nice, catmg   !!!

M + N   = 978

N  =225 + L

M  =225 + J

So

M + N    =  450 + (L + J)

978  =  450  + (L + J)

L + J  = 528

(F + G)   = J

(G + H)   = 225

(H +  I)   =  L

(H + 147)   = L

So

  J       +     225        +      L  =      753  

(F + G)  + (G + H)  + ( H + I)   =   753

(F + G)  +  (H + I)   =   528

(F + G)  + (H + 147)   =528

(F + G)  +  H    =   381

F  + (G + H)  =   381

F + 225   = 381

F  =  156

A +  B   = F

137 + B   =  156

B  = 19

Just as 4M  and Melody found  !!!!

P.S.  - I really liked  this problem ......

Draw a horizontal line through point (2,1) until it is directly below the circle centre. Draw a vertical line from there to the circle centre. You now have a right angled, isosceles triangle with the radius as the hypotenuse (remembering that the hypotenuse here is at 45degrees ). Thus the two legs of this right angled triangle must be of length \(\frac{r}{\sqrt 2}\) by Pythagoras. Add this to the height (1) of the point (2,1).

Thank you!

Hello Guest!

12! = 11! * 12

We can factor 11! + 12! to 11!(1 + 12). 

1+ 12 = 13

11! + 12! = 11!*13

Thus, 13 is the largest prime factor. :)

I hope this helped. :D

=^._.^=

Thanks, catmg.....here's another approach.....

Every minute  the  minute  hand   moves  360/60  =  6°

Every minute  the  hour  hand moves  30°/60  =    (1/2)°

Let  "12"  on the clock face  be = 0°

At  2PM......the  hour  hand  is at  120)*(1/2)  =   60°   and the  minute  hand is at  0°

In 50 minutes after that....the  hour hand  has  moved   (1/2) * (50)  =  25°

And the minute hand has moved   50 * 6 =   300°

So.....the  greater angle  betwen these is   300 - ( 60+ 25)  = 215°

And the lesser angle is    360   - 215    =   145°

For convenience let  Q   = (0, 1) 

Let the center lie at (0,0)

Note that in the equation of the ellipse, "b"    = 1

Then  1/2  the  focal length  =  1/sqrt (3) 

So F1F2  = 2/sqrt (3) =   (2/3)sqrt (3)

We can find "a"   as  sqrt  [ 1^2 + (1/sqrt(3))^2  ]  = sqrt (1 + 1/3)   =  sqrt [ 4/3 ] =  2/sqrt (3)

a^2 =  4/3

A line from Q through R  will have a slope of  tan (-60)  =  -sqrt (3)

So.....the equation of this line is   y  = -sqrt (3)  +  1      (1)

To find the positive x values  of R, sub  (1)  into the equation of the ellipse

x^2 / ( 4/3)  +   [ -sqrt (3)x  + 1]^2  =  1

(3/4)x^2  +  3x^2  -2sqrt (3)x  + 1   = 1

(15/4)x^2  -  2sqrt (3)  x  =  0

x   ( 15/4x -  2sqrt (3)  )  =  0

The second  factor set to 0  will give ys the  x coordinate of R

(15/4)x  - 2sqrt (3)   =  0

x = (4/15)2sqrt (3)

x =  (8/15)sqrt (3) 

And the y coordinate  of  R = -sqrt (3) (8/15)sqrt (3)  + 1  =  -24/15 + 1  =  -9/15   = -3/5

And  QR  = PQ =   sqrt   [ (8/15sqrt (3)^2   +  (1 - -3/5)^2  ]    = 

sqrt[ (8/5)^2  ( 1/3 + 1 ]  =  (8/5)sqrt [ 1 +1/3  ]  =  (8/5) sqrt  ( 4/3)  =  (16/5) sqrt (1/3)  = (16sqrt (3) / 15)  =

(16/15)sqrt (3)

So

PQ   / F1F2  =     (16/15) sqrt (3)    / [ (2/3)sqrt (3)  ]  =    (16/15) (3/2)  =  8/5

Hello Guest!

The important thing to remember when doing clock problems is that the hour hand moves while the minute hand moves. 

At 2:50, our hour hand will be 5/6 the distance from 2-3, and the minute hand will be at 10. 

Between each number is a 5 minute interval, so between the 3 and 10, the difference is 35 minutes, (10-3)*5.

However, since the hour hand will be 5/6 the distance from 2-3, we add 1/6 of the 5 minutes. 

35+1/6*5 = 35 5/6.

So the difference between the minute and hour hand is 35 5/6 minutes, and a clock has 60 minutes. 

Since the clock is 360 degrees, each minute is worth 6 degrees (360/60). 

Thus, our angle is (35+5/6)*6 = 215 degrees. 

However, because we want the smaller angle, we do 360 - 215 = 145. 

145 is the final answer. 

I hope this helped. :)  

=^._.^=

One question only; why is the vertical height from the x-axis to the circle centre \(1+\frac{r}{\sqrt{2}}\) ? Thanks!

As follows:

Speed = distance/time

so

distance = speed *time

Use this formula to find the distance on the way there,

and then again to find the distance on the way back.

Then put the 2 distances equal and solve the arising equation.                                                                                                         

Try it yourself, if you get stuck then show what you have done and ask for more assistance.

Try proofreading your question !

A triangle ABC with three different side lengths had the longest side AC and shortest AB. If the perimeter of ABC is 384 units, what is the greatest possible difference between AC-AB? 

I don't see a word integer(s) in this question.

x^2 - 60x + 851

factor(851) = 23*37

23+37=60

(x+23)(x+37)

a=37 b=23

You can finish it.

The diagram shows 3 identical regular hexagons with a side length .7. If O_{1}, O_{2} O_{3} are their respective centers, what is the area of △O1​O2​O3​?                                                                                    /\

                                                                                              is that  7  or  .7 ???

Find all values of x such that x/(x + 4) = -8/(x + 7). If you find more than one value, then list your solutions in increasing order, separated by commas.

Hello Guest!

\(\color{BrickRed}\frac{x}{x+4}=\frac{-8}{x+7}\\ x^2+7x=-8x-32\\ x^2+15x+32=0 \)

         p          q

\(x=-\frac{p}{2}\pm \sqrt{(\frac{p}{2})^2-q}\\ x=-7.5\pm\sqrt{7.5^2-32}\\ x=-7.5\pm \sqrt{\frac{97}{4}}\\ x=-7.5\pm 0.5\sqrt{97}\\ x=-7,5\pm 4.9244\)

\( x\in\{-12.42443,-2.57557\}\)

  !

Sorry your answer was incorrect, I have no idea about your solution but thanks for the effort