All Answers 358089 Answers

How does the air temperature have anything to do with the distance of water below the person? Please explain. In addition, please message to Melody or other moderators instead because I will be going away.

Thanks

GLiu

what is the smallest number with 13 factors?

Answer? = 2^13=8,192

We need to solve for the time to hit the ground...and we have

0  = -4.9t^2 + 15  subtract 15 from both sides

-15 = -4.9 t^2    divide both sides by -4.9

-15 / - 4.9  = t^2       take the square root of both sides

t = about 1.75s

So......the soccer ball landed

12.3 (1.75)  = about 21.525m from the base of the cliff

Yep....that one is good, too....

Yea that is pretty cool Chris - DS would have to see it it worked as a little icon. 

That is just trial and error.

Here is another but it is not as coold as chris's.

MBA Test Question? How many marks would I get if I got right??????!!!!!

1st. part- Dr. Taylor should have saved this amount=$850,103.36.

2nd. part- Dr. Taylor should have saved this amount=$580,544.54.

3rd. part- Dr. Taylor should have saved this amount=$84,473.95.

The grand total of the above three sums comes to=$1,515,121.85

This amount will enable Dr. Taylor to withdraw an annual annuity payment of $118,523.01 every year for 25 years. So, the answer is certainly "Yes"!.

The formulae used to calculate these amounts are in your book!. If you don't know which ones to use, then I would say, you are in trouble!!!.

That was a good one CPhill. XD.

How about this one, DS???

Yes, prime.....!!!

For the forum, or the calculator???

Or a dead dragon...

We need to find the time the ball traveled

We can use this equation :

0 = 3.20sin(55)t  - 4.9t^2   divide through by t

0 = 3.20sin(55)  - 4.9t    add 4.9t to both sides

4.9t = 3.20sin(55)    divide both sides by 4.9

t = 3.20sin(55)/ 4.9  = about .535s

So.....the horizontal displacement  is given by :

d  = 3.2cos(55)(.535) = about  0.982 m

Find the following limit:
lim_(x->4) (x sqrt(x)-8)/(x-4)

(x sqrt(x)-8)/(x-4) = (x^(3/2)-8)/(x-4):
lim_(x->4) (x^(3/2)-8)/(x-4)

Factor the numerator and denominator:
lim_(x->4) ((sqrt(x)-2) (4+2 sqrt(x)+x))/((sqrt(x)-2) (2+sqrt(x)))

Cancel terms, assuming sqrt(x)-2!=0:
lim_(x->4) (4+2 sqrt(x)+x)/(2+sqrt(x))

lim_(x->4) (4+2 sqrt(x)+x)/(2+sqrt(x))  =  (4+2 sqrt(4)+4)/(2+sqrt(4))  =  3:
Answer: | =3
 

Change it to a dragon ... 

... Change it...

Why did u give me an pic, and email...??? If I'm supposed to email u, u can forget it:)

I believe we can use similar triangles to solve this

In one second the current rate is 29/36m/s   and the swimmer's rate is 20/36m/s...so...

29/20  = d/500   where d is the distance downstream ....multiply both sides by 500

500*29/20 = d =  725m

What r u talking about?!?!?!?

Still waiting 

O.K.  

U an call me, if U like.

d.

(the Guest by U)  ►http://cdn01.trixum.de/upload2/26200/26120/4/4420dc98321dd580a019e3dbfe98ca209.jpg

Sorry, I was eating dinner... My name is Victoria, what is yours?

Still waiting 

Hi, how can I call U - boared Lady ??

cette@web.de

Math you shall learn, Young Hayley

17 years of age, you are.

Get this wrong, how did you?

~Yoda

Vf = Vi + (-9.8m/s^2)*t  whereVf is the velocity at any time, Vi = 16.67m/s and t = 1  ......so we have

Vf = 16.67/m/s + (-9.8m/s^2) (1s)  =

Vf = 16.67m/s - 9.8m/s  =

Vf = 6.87m/s   [ the potato is still rising after 1 sec ]

Initial Velocity =  16.67 m/s

Accel of Gravity = -9.8 m/s^2

vf = vo + at      Velocity final = velocity original + accel x time

vf = 16.67  + (-9.8)(1)

   = 16.67 - 9.8 = 6.87 m/s