All Answers 358177 Answers

sorry, i'm at the point of combining like-terms and it's

x^2

-x

____

If Angle A is 30º and side b is 6 feet, find a

Assuming a and b are legs of a right triangle........

We have a 30 -60- 90 trianlgle      and the sides a, b and c  are in the respective ratio  1 : √3 :  2

So...side  a  =   6 / √3    = about 3.464 

It's just part of a rational equation i am doing and i was wondering if it would just turn out to be zero or just x

That does NOT 'cancel out' ....can yo give mor information?

x^2 - x    =  x (x-1)

y=(2x^2-5x-3)/(x^2-4x+3)     factor  numerator/denominator

y = [ ( 2x + 1) (x - 3)] / [ (x - 1) (x - 3)]      "cancel" the (x - 3) factor and we have

y =  [ 2x + 1] / [ x - 1 ]

Here's the graph : https://www.desmos.com/calculator/yc7cj5ahco

There is a horizontal asymptote at y  = 2

A vertical asymptote occurs at x = 1

Domain =  all x  except x = 1  and x = 3

Range  = all y except y = 2

y intercept occurs at  (0, -1)

x intercept occurs at ( -1/2, 0)

Thank You So Much:) my nest quetsion consist of: A new SpaceX rocket is being tested and launches from the ground. The first stage rocket engine provides constant upward acceleration during the burn phase. After the first stage engine burns out, the rocket has risen to 105.0 m and acquired an upward velocity of  The second stage fails to fire. The rocket continues to rise, reaches maximum height, and falls back to the ground. Neglect air resistance. The time interval during which the rocket engine provided the upward acceleration, is closest to

3.0 s.

4.2 s.

1.5 s.

1.7 s.

4.0 s.

What kind of equation?

Thanks! ; )

=   -8/2187    x   sqrt 2   =    simplest form without converting to decimal

We'll use a couple of different equations regading position and velocity

First:  v = vo + at       vo= 50  a = -9.8 m/s^2    FInd the point where the rocket reaches its apex, v=0

0 = 50 -9.8t      results in  t = 5.1 deconds

How high did it go?

x = xo + vo t + 1/2 a t^2       xo= 105 m    vo = 50 m/s    and   a = -9.8 m/s^2  

   = 105  +  50 (5.1)  - 1/2(9.8) (5.1)^2   = 360-127.45 = 232.55 m

Two solutions satisfy p:

p ≈ 1.5155087790640846000579494

p ≈ 3.1415926508258620347077231

P.S. notice the second solution is an approx. of Pi !!!!!!!!.

Solve by elimination:

x + y + z  =  10

-x + y - z  =  0

-x + z  =  1

Combine the first two equations by addition (adding down the columns):

     x + y + z  =  10

    -x + y - z  =    0

    --------------------

         2y       =  10     --->     y  =  5

Replace y with 5 in the first equation:    

     x + y + z  =  10     --->      x + 5 + z  =  10     --->     x + z  =  5

Combine this equation with the third of the original equations by addition:

     x + z  =  5

   -x + z  =   1

   ---------------

         2z  =   6

           z  =  3

Since  y = 5  and  z = 5.5,  substitute these values into the first equation:

     x + y + z  =  10     --->      x + 5 + 3 =  10     --->     x + 8  =  10     --->      x  =  2

I will assume you mean this:   -8 sqrt(2187)

= -8 sqrt (81*27) = -8 (9) sqrt(27) = -72 sqrt (27) = -72 sqrt(9*3) = -72(3) sqrt (3) = - 216 sqrt3

The distance equals the initial 50 miles at 75 mph + additional miles at 45mph.

50 m / 75mph = 2/3 hr = 40 minutes

The REST of the time was at 45 MPH    2 hr 15 min  - 40 min = 1 hr 35 min   @ 45 MPH

Rate times time = distance        45 MPH x 1 hr 35 min = 45 mph x 1.58333hr = 71.25 miles

PLUS the initial 50 miles = 121.25 miles    

Is this a real question? I'm sorry I seem rude w/ this question.

SEE YOUR ANSWER UNDER YOUR SECOND POST!!!.

Suppose you're driving to the city at a constant speed of 75 mi/h for 50 miles. Then construction slows you to a pokey rate of 45 mi/h for the remainder of the trip. You arrive after driving 2 hours and 15 minutes. How far do you live from the city? A. 180 mi B.120 mi C.200 mi D.71 mi 71 mi.

You can cover the first 50 miles in: 50/75 x 60 =40 minutes.

Since total time taken to drive is 2 hours and 15 minutes=135 minutes, then

135 - 40=95 minutes or 1 hours and 35 minutes that you were driving @ 45mph, therefore:

95/60 x 45 =71.25 miles that you drove @ 45 mph, therefore:

71.25 + 50 =121.25 miles-That is how far you live from the city, or "B" roughly!.

yes it can.

(2x + 1)(x-3)

hint

The way to check if a quadratic will factorise into whole numbers is to look at  b^2  - 4ac,if that's a  square number then the quadratic will factorise.

In the case of your problem, b^2  - 4ac   = 1 which is a square.

the nth term is    a  + (n-1)d   where a is 1st term and d is common difference.

so  tenth term is                     -5    +  {  (9)  x  (- 1/2)  }  

sure you can do the rest.

x = x0 + vot + 1/2 a t^2         In this instance x0 = original position = 0   vo = original velocity = 0

so you are left with  x = 1/2 a t^2  = 1/2 (32) (1.4)^2 = 16 ( 1.96) = 31.36 ft deep  (using 'a' = 32)

is that right

Muscles! :D

Divie by 3 to get   3:4:7

Least Common Multiple ????   966,000

What Are The Numbers For?