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I don't know

W2  and !9    ⇒   "B"

complete the square help needed thank you

The yearly loss is

.005 * 1,340,000 * 365    =  2,445,500    =  "a"

Multiplying polynomials help 

The diagonal of the square must be  <  9  cm  for the square to fit

The diagonal is given  by

diagonal  =     sqrt  ( side^2  + side^2)    = sqrt  [ 5^2 + 5^2 ] = sqrt  [ 50]  ≈  7.07  cm

So..the square will fit inside the circle

Total Income  =

2300 + 640   =   2940   (1)

Total  Expenses  =

775  + 308  +  385 + 273  + 87  + 30    = 1858   (2)

Money left  =    (1)  - (2)  =   2940   - 1858    = $1082

See the following  image :

AZDY  forms a  kite with angles AZD, AYD  = 90    

And  ZAY  = 40

So...angle ZDY =    360 - 2(90)  - 40  = 140

And in triangle ZDY, because DZ  = DY....then triangle ZDY is isosceles with angle YZD  = [ 180 - 140 ] / 2  = 

40/2   = 20

And CXDZ  also forms a kite with angles DXC, DZC  =  90

And angle ZCX  =  80

So....angle ZDX  =  360  - 2(90) - 80  =  100

And in triangle ZDX, because DX  = DZ....then triangle ZDX is isosceles with angle DZX  = [ 180 - 100 ] / 2  =

80/2   =  40

And angle YZX   = angle YZD  + angle DZX  =  20 + 40    =  60

v=0 when car stops

v=vo + at       vo= 20 m/s    a=-5 m/s^2

0=20 +(-5)t

t= 4 seconds

x=xo + vot + 1/2 at^2

x = 0 + 20(4) + 1/2 (-5)(4)^2

    =  80 + (-40)= 40 m       (as Chris found) 

What is the anwser

??????????

When A is reflected over the x axis to B....the new point is  ( 3, -4)

Call this the base of the triangle AB  = 8  units

And the equation  of the line on which this base lies is  x  = 3

When B is reflected over the line y  = x  the new point C has the  coordinates  ( -4, 3)

The  distance from this point to the line  x  =  3  is the height of the triangle  = 7

So....the area  of ABC   = (1/2) AB * 7   = (1/2) (8)(7)    = 28 units^2

When you say " the sum of the squares of each number is 153", I take to mean "the sum of the squares of both numbers is 153". Based on that assumption, then we have:

x - y =9................................(1)

x^2 + y^2 = 153..................(2)

From (1) above: x =9 + y.   sub this into (2) above:

(9 + y)^2 + y^2 = 153, solve for y

y^2 + 18y + 81 + y^2 =153

2y^2 + 18y + 81 - 153 = 0

2y^2 + 18y - 72 = 0 factor:

2(y - 3)(y + 12) = 0   divide both sides by 2

(y - 3)(y+ 12) = 0

y = 3     or    y = -12 and:

x =12    or    x =- 3

x*y =3*12 =36  or -3*-12 =36

Thank you so much!

Not great at Physics, but I'll give it a shot

vf^2  = vi^2 + 2ad

vi  = initial velocity  = 0....so we have

(50m/s)^2  =  0    + 2 (4m/s^2) * d

2500m^2 / s^2  =   8 m/s^2  * d          divide both sides by  8m/s^2

[2500m^2/ s*2]  / [ 8m /s^2]   = d   =  312.5 m  ......correct  !!!

Second one

vi  = 20m/s

vf  = 0   [ we are stopped at the end ]

a  = -5m/s^2   [ we are slowing down ]

So

0  =  [20m/s]^2  + 2(-5m/s^2) * d

0 =  400m^2 / s^2     - 10 m/s^2 * d       multiply through by  s^2

0  = 400m^2  - 10m  * d

10m * d  =  400m^2        divide both sides by 10m

d  = 400m^2  / 10m   =  40 m  = stopping distance

Thanks :)

x=xo+ vot + 1/2 at^2       xo = 0    vo=0    a = 4

v=vo + at     v=50 m/s   vo=0    a = 4 m/s^2        solve for t =  12.5 s

x = 0 + 0 + 1/2 (4)(12.5)^2  =  312.5 m     You got it! (Well.....at least we both arrived at the same answer!)

Two vertices of an obtuse triangle are (6,4) and (0,0). The third vertex is located on the negative branch of the x-axis. What are the coordinates of the third vertex if the area of the triangle is 30 square units?

The height of the triangle is  4

So

Area of  triangle  =  (1/2) base * height

30  = (1/2) base * 4      simplify

30  = base * 2        divide both sides by 2

15  = base

The third vertex is   ( -15,  0)

Since AB  and CD are parallel

Then angles A  and D are supplemental  ....so

A  + D   = 180

2D + D   = 180

3D  = 180

D   = 60

And  A   = 2D   = 2(60)   = 120°

Sorry...I posted incorrect answer....

     Agree with Chris' answer....

We have either

y  = 2x   or   x  =  2y

Look at their graphs, here : https://www.desmos.com/calculator/t2gtfjbzrd

So....they split the plane into 4 regions

x^2 - 5x + 4  = 0

Factor  the left side as  :

(x - 1) ( x - 4)  = 0

Set both factors to 0   and solve for x and we have that   x  = 1, 4

The ball is thrown HORIZONTALLY, so the initial vertical velocity = 0 m/s   and the initial horizontal velocity = 25 m/s.

1 The ball's 'x' position is given by    x = x0 + v0t           x0=0    vo=25   (there is no 'a' in the x direction)

2 The ball's  'y' position is   y = yo + vot  + 1/2 a t^2      where yo=63  vo=0  a = -9.8 m/s^2

Use 2 to find t when y = 0  (when the ball hits the ground)

    t = 3.59 s

Use this time to find the x distance the ball travels before hitting the ground (equation 1)

    x = 25(3.59) = 89.64 m

3 Velocity when it hits the ground will be BOTH an x velocity and a y velocity.

     the x velocity is given as  25 m/s

     the y velocity is given by  vy = vo + at       v0= 0    a = -9.8 m/s^2        t = 3.59 s

      vy = -9.8(3.59) = -35.18 m/s

     'VECTORIALLY' add x and y together to get the velocity MAGNITUDE    sqrt(25^2 + 35.18^2)= 43.16 m/s

I will take a crack at this one:

Will consider this in 2 separate steps:

1 - Distributing 3 identical apples among 4 friends:

(3+ 4 -1) C (4 - 1) =6C3 =20 different ways.

2-Distributing 3 identical oranges among 3 friends:

(3 + 3 -1) C (3 - 1) =5C2 =10 different ways.

So, the total =20 + 10 = 30 different ways.

Note: somebody should check this.

Got it! Thanks ^-^

(4c+1)(16c^2-4c+1)

n can be any multiple of 6 so the sum is infinite.  

This answer is compeletely  wrong so skip it.