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I interpret the question as: "How far does he go before he stops?"

In this case the appropriate equation is v² = u² + 2as  where v is final velocity (0), u is initial velocity (8.1m/sec), a is acceleration and s is distance.

The acceleration here is -9.8*sin(20°) m/sec², the component of gravitational acceleration acting parallel to the slope (negative because it is retarding the skier).

0 = 8.1² - 2*9.8*sin(20°)*s

s = 8.1²/(2*9.8*sin(20°)) metres

$${\mathtt{s}} = {\frac{{{\mathtt{8.1}}}^{{\mathtt{2}}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{9.8}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{sin}}{\left({\mathtt{20}}^\circ\right)}\right)}} \Rightarrow {\mathtt{s}} = {\mathtt{9.787\: \!286\: \!055\: \!842\: \!569\: \!1}}$$

s ≈ 9.8 metres

Oh! Thx! I didn't see that!  :)

Thank you, guys! :)

We can use the Law of Cosines

17^2 = 11^2 + 14^2 - 2(11)(14)cos(L)

289 = 121 + 196 - 308cos(L)

-28 = -308cos(L)        divide both sides by -308

(28/308) = cos(L)        take the cosine invese

cos^-1 (28/308) = about 84.8°

I'm sorry, I can't understand what you are saying.

You dived 11,17,14 or you can muxtiyply 11, 17 then what you get for

that with the 14 then dived, what you got for 11,17 then 14 and what you got

from 11,17. 

We can use the Law of Cosines here

a^2 = 7^2 + 9^2 - 2(7)(9)cos(46°)

a^2 = 49 + 81 - 126(.695)

a^2 = 42.43        ......take the square root of both sides

a = about 6.5

Since the triangle is a SAS situation, use the Law of Cosines:  b = 9   c = 7

a² = b² + c² - 2·b·c·cosA

a² = 9² + 7² - 2·9·7·cos46

a² = 42.473

a  =  6.5

g(t)=t(4-t)^1/2        we can use the product and chain rules here

g'(t)  = (4-t)^1/2  + (1/2)t(4-t)^(-1/2)(-1) = (4-t)^1/2 - (1/2)t(4-t)^(-1/2) =

.......(factoring).....   (4-t)^(-1/2) [ (4-t) - (1/2)t]   =  (4-t)^(-1/2)[4- (3/2)t ] =

(4-t)^(-1/2) [(8 -3t)/2]

A good equation to know is the  point-slope  equation, used when you know a point and a slope.

y - y1  =  m · (x - x1)           where m is the slope and (x1, y1) is the point.

For your problem,  m = 3  x1 = 5  and  y1 = 2:

y - 2  =  3(x - 5)

Now, it depends in what form you want your answer:

To get it into y-form:

y - 2  =  3x - 15

y  =  3x - 13

The prefix "milli" means "one one-thousandths", so to covert meters to millimeters multiply by 1000.

Here is one of the best pages on the internet that covers Phi (and it's reciprocal, phi):

-y > 2/3

MUutiply both sides by -1:  (If you divide or multiply both sides of an inequality by a negative number, you                                                must change the direction of the inequality.)

-y · -1  <  2/3 · -1

y  <  -2/3

First, you will need to find the distance from A to C, which is side b.

Since you have a SAS triangle, use the Law of Cosines:

b² = a² + c² -2·a·c·cosB

b² = 63² + 91² -2·63·91·cos153.4

b² = 22502

b  =  150 mi

150 miles at 300 miles per hour  =  1/2 hour

2f² - 9f + 5  =  0

Quadratic formula:  x  =  ( -b ± √(b² - 4ac) ) / (2a)

In this case:  a  =  2         b  =  -9         c = 5

x =  ( --9 ± √( (-9)² - 4(2)(5) ) ) / ( 2(2) )

x =  ( 9 ± √( 81 - 40 ) ) / ( 4 )

x =  ( 9 ± √( 41 ) ) / 4 

You, do 91mi  dived as 63mi then X what you got when you dived with 153.4

Chris said all the integers were different!  

look here.  

Can you use each digit only once, or can you use some of the digits more than once?

Must you use each digit?

Can you use addition, subtraction, multiplication, division, exponentiation, finding roots, factorials, and including negative signs?

In summary, what are all the rules?

Since  12 - 6  =  6:

     6 = Cloud

If you have specific questions, ask.

Can you be more specific?

You are looking for √6,000,669.

You can use the calculator at this site to find a good approximation of this value.

I'm far more into math, then I am into physics, but I'll take a shot at this:

the formula for location is:  d  = -4.9t² + vt

If the velocity is  8.1  along the slope, the vertical component of this will be found by using sin:

    sin(20°)  =  v/8.1   --->   v = 2.770

The formula becomes:  d  = -4.9t² + 2.77t

Since this formula describes a parabola, the maximum point (the stopage point) is halfway between where the distance is 0 (when it starts) and where it again becomes 0.

Solve:  0  = -4.9t² + 2.77t     --->     0  =  t(-4.9t + 2.77)     --->  t = 0  and  t =  0.5653

Halfway between 0  and  0.5653  is  0.28 seconds.

Again, no guarantee.