- (x+y) = (-1)(x+y) = -x - y

12/6+28 = 30

Because 12 divided by 6 =2. Then, 2 +28=30.

Given a = -5 ft/sec^2

and xf = 45     and   t = 1        find vo

xf = xo + vot + 1/2 a t^2          xo = 0

45 = vo (1) + 1/2 (-5) 1^2       yields vo = 47.5 ft/s

Find TIME to stop  given  vo = 47.5 ft / s      a = -5 ft/s^2    and vf=0 (the car has stopped)

vf = vo + a t

0 = 47.5 - 5 t    yields  t = 9.5 s

Find DISTANCE to stop given  vo=47.5 ft/s  a = -5 ft/s^2   and  t= 9.5 s

xf=xo + vo t + 1/2 a t^2        xo=initial position being measured from = 0

xf = 47.5 (9.5) + 1/2 (-5)(9.5^2)    Yields   xf = 225.625 ft  to stop

First, the volume of the cube has to be in cm^3   not cm^2

Second.....there needs to be a diagram or something....missing?

what is the area of what?

You're welcome, of course!

thank  you

Divide them to find out

5.7 x 10^9  / 3 x 10^8   =  1.9 x 10^1 = 19 times larger

Graph it to find vertex = -4 , -3      (desmos calculator or such)

-b/2a = -4    Substitute this value of 'x' to find y = - 3

Derivative = 2x+8 = 0    x = -4

i know thank u 

its 4 

gg m8 

2(n-3) <= 10

n-3 <= 5

n <= 8

In loose terms, it is where you prove an 'assumption' in mathematical (geometrical, trigonomic, algebraic) terms using axioms (truths) and laws.

No table is shown.......

P (two orange in a row without replacement ) =  2/13 * 1/12 =   2 / 156   =   1 / 78

We have the identity that :

cosh²(x) - sinh²(x) = 1    

cosh²(x)  - 4^2  = 1 

cosh²(x)   = 1 + 4^2

cosh²(x)  = 17

cosh(x)  = ± √(17)

And 

tanh(x) = sinh(x)/cosh(x)    ....so......

tanh(x)  =  ± 4/ √(17)

Generally a cm^3 is considered to be equal to  a   ml

sooooo    1500 ml / 250ml  = 6 of the 250 ml containers.

No problem.

o.o um.. Thanks! lol

6 / √21      multiply top/bottom by √21

[ 6 √21 ] / [√21 √21 ]  =

[6 √21 ] / 21   =

(6/21) √21 =

(2/7) √ 21  =

 [ 2 √ 21 ] / 7

6 - e +de-2e = 6 + de - 3e = 6 + e (d-3)

250  x  10/100% = 25

How do you knw what my tan 64   IS???     haha

tan 64 DEGREES = 2.050

tan 64 RADIANS  =  2.349