All Answers 352963 Answers

Call the height of the triangle, y

Call the base length, x

And

sin ( theta)  =   y / hypotenuse   ⇒  hypotenuse * sin (theta)  =  y

cos (theta)  =  x / hypotenuse  ⇒ hypotenuse *cos (theta)  = x

So.....this must mean that the hypotenuse  = 10

Here's an graphical way of figuring this out

C (n , 4)   =  n! /  ( [ n - 4] !  * 4! )   .....so....we require that.......

C ( n , 4 )  ≥  365

n! /  ( [ n - 4] !  * 4! )  ≥  365

n! / [ n - 4]!  ≥   4! * 365

n! / [ n - 4]!  ≥  8760 

n * (n - 1) (n - 2) (n - 3)  ≥  8760

Look  at the graph here : 

https://www.desmos.com/calculator/j9rafrlbpn

Note that  the minimum n that makes this true ≈  11.3

So.....we need 12 friends

Proof  C(11, 4)  = 330    too few groups of 4

But  C( 12. 4)  = 495     .......we even have a few "leftover" groups of 4  !!!!!! 

why??

why??

27/3=9 wich means 9$ per hour

32/4=8 wich means 8$ per hour

Anne's deal is the best

I agree i believe the answer is J

I think it's J

Okay so this is a simple multiplication problem (for the most part). 

If a runner is running at 2.4 m/s what is the distance it will travel in lets say 2 minutes.    <----Second time interval

Well first you'd want to convert those minutes into seconds. So there are 120 seconds in 2 minutes.

Now into the multiplication,             if for every second the runner is running 2.4 meters then you would multpily 2.4 by however many seconds

(2.4) x 120 = 288 meters 

Thanks, heureka....I am very intersted in this problem....your answer includes some combinatorial identities........as I am not too familiar with these, could you show me how these two are derived ??

[ n / ( k + 1) ]  =  [  (n - k ) / ( k + 1) ]   *  [ n / k ]

And

[ n / (k - 1) ]  =  [ k / (n - k + 1) ] * [ n / k ]

Also......what allows us to do this ??

2 [ (n - k) / (k + 1) ]  =  1

n - k   =  [ k + 1 ] / 2

Thanks....!!!!

Only things without mass can reach the speed of light. One or both of Einstein's Theories of Relativity describes why nothing can go faster than the speed of light. I think of it as a time paradox - once you reach the speed of light, you stop feeling time.

Same thing bro

Pretty neat, helperid  !!!!!

Given that x^2 +ax +b =(x+2)^2 -9 work out a and b

Just expand (x+2)^2 -9  = 

x^2 + 4x + 4 - 9  =  x^2 + 4x - 5    ⇒ a = 4      b = -5

 x^2 - 10x - 5

Take  1/2 of 10  = 5.....square it  = 25 .... add and subtract it

x^2 - 10x + 25 -  25           factor the first three terms

(x - 5)^2  - 25     =  ( x + (-5) )^2   -  25    ⇒    p =  -5        q  = -25   

write expression 2x^2 - 8x + 19 in form a(x+b)^2 + c

Factor out  2

2 [  x^2  - 4x  +  19/2]

Take 1/2 of 4  = 2......square it......= 4.....add and subtract it

2 [  x ^2 -  4x  + 4   +  19/2  -  4 ]

2 [  x^2 - 4x  +  4  +  19/2  - 8/2 ]      factor the first three terms

2 [ ( x - 2)^2  +  11/2] 

2(x - 2)^2  +  11

The minimum point on the graph  is   (2,11)

Since this minimum lies above the x axis......the graph never crosses that axis

It includes user-input concentric circles and another diamond pattern, each in separate folders.

p^2 -5p - 14 

We are looking for two integers that sum to -5  and multiply to -14

The two that we need are     -7   and  2

So........we write

(x - 7)  (x + 2)   ......and that's it  !!!!

Thanks, geno........here's another way

Tne x value that minimizes the function is given by  

-b / [2a ]       where  a = 3  and  b = -15     .....so we have

- [ -15] / [  2* 3 ]  =   15/6   =  5/2

So......putting this into the function we have

3 (5/2)^2 - 15 (5/2)  +   k  = 1 

75/4 -75/2 + k  = 1

-75/4  + k =  1

k =  79/4   = 19 + 3/4

you can see how you got 58

I'm not good at russian but I think I know what answer you're searching.

0,29(x+200)-0,89x=216

0,29x+58-0,89x=216

0,29x-0,89x=158

-0,6x=158

-0,6x/-0,6=158/-0,6≈-263,3

The speed of light (299 792 458 m/s in vacuum) is the fastest we know. It is not possible to smoothly accelerate a material object to, or beyond, the speed of light. 

If you're having an organization problem, use a folder for the papers of the current unit (one side for HW, the other for notes), and use tabs in a folder for all previous units. If you need time management help, do as much as you can in that class, and do the rest in othe classes (if possible) and/or on the bus (if you take the bus to school).

To find the value of s for which:  sec(s)  =  1.0806.

Since  sec(s)  =  1/cos(s):     sec(s)  =  1.0806.    --->     1/cos(s)  =  1.0806

                        Multiply both sides by cos(s)                               1  =  1.0806 x cos(s)

                        Divide both sides by 1.0806                    1/1.0806  =  cos(s)

                         Simplifying                                                 cos(s)  =  0.9254

Now, use your calculator to find the inverse cos of 0.9254.

[Inverse cos  is  arc cos  is  2nd cos]

Let the base of triangle ABC  be  " x ".

Now, let's draw a height from side  x  to angle  C .

This makes a right triangle with the hypotenuse of  y .

sin 70°  =  opposite / hypotenuse

sin 70°  =  height / y

y sin 70°  =   height

area of triangle ABC  =  (1/2)(base)(height)

area of triangle ABC  =  (1/2)(x)(y sin 70°)       This equals 30 because the problem says so.

Now let's look at triangle  PQR. Let the base be  x  .

This makes its height  =  y sin 110°

And the area of PQR  =  (1/2)(x)(y sin 110°)

Let's say

30  =  (1/2)(x)(y sin 70°)

a  =  (1/2)(x)(y sin 110°)        We want to know what  a  equals.

30 / a  =  [ (1/2)(x)(y sin 70°) ] / [ (1/2)(x)(y sin 110°) ]

30 / a  =  sin 70° / sin 110°

30 / a  =  1

30  =  a