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nce the line of symmetry is x = 2, the x coordinate of the vertex  must = 2

And the x coordinate of the vertex is given by :   -b/2a.......so we have that

2  = -b/[2a]      →   -4a = b

And we have these  equations

1 =  a(1)^2   - 4a(1)  + c    →   1 =  -3a + c       (1)

-1 = a(4)^2   - 4a(4)  + c    →  -1 =  c      (2)

Subbing (2)  into (1)  we have that    1 =  -3a - 1   →  2 = -3a →  a = -2/3

Which means that b  = -4(-2/3)  = 8/3

So....the equation of the quadratic is :

y = (-2/3)x^2 + (8/3)x  - 1

And we know that

0 = (-2/3)(√n +2)^2  + (8/3)(√n + 2)  - 1          let √n + 2   = m.......and we have that

0  = (-2/3)m^2 + (8/3)m  - 1      multiply both sides by 3

0 = -2m^2  + 8m - 3     complete the square on m

3 - 8 = -2 (m^2 - 4m + 4)

-5 = -2 (m - 2)^2

5/2  = (m - 2)^2       take pos/neg square roots

±√[5/2]  = m - 2

±√[5/2] + 2  = m     back-substituting, we have

±√[5/2] + 2 = √n + 2   and since we want the larger root

√[5/2]   = √n

n = 5/2  = 2.5

Here's the graph :  https://www.desmos.com/calculator/102umpw19h

The larger root = ( √2.5  + 2 ,  0 )  ≈  ( 3.581, 0 )

In a right triangle    the Sin of the  angle is the opposite side over the hypotenuse  =  12/37

20 km / (r+2)   + 1/3  =   20km/ (r)                     (1/3 hr = 20 minutes)

20 (r)  + 1/3( r)(r+2) = 20 r +40

20 r   + 1/3 r^2 + 2/3 r  = 20 r + 40

1/3 r^2 + 2/3 r -40 = 0                              quadratic formula shows r = 10      r+2 = 12

5 songs 

and

3 games

5 songs = $2 x 5 = $10

3 games = $5 x 3 = $15

He downloaded 8 items. Add 5 and 3, you get 8.

He has $25 dollars on his gift card. Add 10 and 15, you get $25.

I'm sorry I tried but I only got that a0 is 1

In the old days it was possible to recieve up to 5 hearts from each person. But still that barely makes a difference as 111426 hearts is still a massive number to achieve regardless.

I'm curious too.  

The answer is b

c = 14*cos(38º) 

Since it starts in the upper left-hand corner of the graph and ends in the upper right-hand corner

of the graph it must be both an even degree polynomial and the coefficeint of the even degree must be

positive.

Therefore, the answer is ........

Since the ratio of team assistants to player is 1/6, there are 1/6^th as many assistants as players.

(1/6) · (36) = 6     --->     there are 6 team assistants.

The ratio of coaches to assistants is  3/6  =  1/2.

C = 85.00 + 0.48·d

Draw altitude AP with P on BC.

Since B = arcsin(24/25), sides AP and AB ( of triangle(APB) ) must be in the ratio 24/25.

Since C = arcsin(12/13), it is also true that C = arcsin(24/26) and that sides AP and AC

( of triangle(APB) ) must be in the ratio 24/26.

In triangle(APB), if AP = 24 and AB = 25, then BP = 7 (Pythagorean Theorem).

In triangle(APC), if AP = 24 and AC = 26, then CP = 10.

Since BP + CP = BC and since 7 + 10 = 17, BP must equal 10 and CP must equal 7.

The area of triangle(ABC) = ½·base·height =  ½·BC·PA = ½·17·24 = .........

c = children tix                                             value = 12c

a = adult tix =  3c                                         value =  18(3c) = 54c

s = student tix = 500-c-3c = 500-4c             value = 15 (500-4c)                              summed = 8040

12c   + 54c  + 15 (500-4c)    = 8040     solve for   c     

                                                                   then   a = 3c      and   s = 500- 4c

angle(A) = 46.8.

sin( angle(A) ) = a / c   --->   sin( 46.8 ) = 3.2 / c   --->   c  =  3.2 / sin( 46.8 )  =  4.4

AB = 4.4.

AC can't be larger than AB, so ...

The two triangles, EBQ and CDQ are similar by A-A-A.

Thus, the areas of the triangles are in the ratio of the squares of the corresponding sides.

Since 3BE = 2DC, BE = (2/3)DC and the ratio of the areas of triangle(EBQ) and triangle(CDQ)

is 4/9.

Since triangle(QDC) has area 36, triangle(EBQ) has area (4/9)·36  =  16.

Assuming that P, Q, and R are on BC (in the order B, P, Q, R, C):

Angle(BAC) = 90.

Since AQ is an angle bisector, angle(BAQ) = 45.

Since P is between B and Q and angle(PAQ) = 13, angle(BAP) = angle(BAQ) - angle(PAQ)

              --->   angle(BAP) = 45 - 13 = 32.

Since AP is an altitude, angle(BPA) = 90.

In triangle(BAP), angle(BAP) = 32 and angle(BPA) = 90   --->   angle(ABP) = 58.

In triangle(BAC), angle(BAC) = 90 and angle(ABP) = 58   --->   angle(ACB) = 32.

Since AR is a median of a right triangle, RA = RB = RC.

Since RA = RC, triangle(ARC) is an isosceles triangle.

Since angle(ACB) = 32, ange(CAR) = 32.

The angle right next to 126   must add   with 126   to 180 degrees (straight line )

then the three angles inside the triangle must add to 180 degrees     allowing you to find 'x'

It takes lots of time and lots of answering questions. 

He is like one of the Founding Fathers...been here for years ,...... and answers a L-O-T of Q's....

   he's a math guru....and a Moderator for the website.

E_m * E_p  = k/4pi

1/2 E_m  * 2E_p  = k/4pi

so halving the momentum error  DOUBLES the position error   ....   a 100% increase

I'll give you the answer but you have to solve it.

∠BAP = ∠ACB = ∠RAC = arctan( AB / AC )