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I don't know

7-4+3x0= 1

100 * (1 + 20%) * (1 + 10%) =

100 * ( 1.20) * (1.10)   =

100 (1.32)  =

132

100 * (1 + 20%) * (1 + 10%)

100* 1.20* 1.10 =132

Yes I understood that  :)

I don't think the other one was there when I started answering.

It is really annoying when people post their question twice!   

An appology would be nice from you guest ! 

Sorry, Melody....I didn't see your answer here.....I answered the one posted above.....fortunately, our answers agree.....LOL!!!!!

Anchor Bush and Putin in their seats.....but remember ....we can seat them in 2 ways.....Bush to Putin's left or vice-versa....

Any of 5 remaining people can be chosen to sit between them   = 5  ways

And the remaining people can be seated in  4! ways = 24 ways

So  2 * 5 * 24   =  240 arrangements

tan^-1(x) represents the angle that has x for a tangent.

For example, tan^-1(1.732...) = 60^o or any addition (or subtraction) of 180^o of that angle; so, answers of 240^o, 360^o, etc; also, answers of -120^o, -300^o, etc. are also correct.

 Seven people including Bush and Putin are to be seated at a round table. In how many ways can the seating plan be arranged such that there is exactly one person who sits between Bush and Putin?

Seat bush and Putin first.

These problems always have the table floaing in space so any one chair is equaivalent to any other. 

so

there are 2 ways that Bush and Putin can be seated.

Now there is a choice of 5 people that can be seated between them.

so that is 2*5 ways so far.

Now there are 4 more people to sit that is  4! ways

so altogether there is       2*5*4! = 10*24 = 240 ways 

If there is no difference between looking at the picture in a clockwise manner or an anti-clockwise manner, there are 5 persons who can be the individual placed between B and P. Of the four remaining, any of four can be placed next to B, any of three can be placed next to that person, any of two can be placed next to that person, and there is one person remaining to be placed next to P. (There is no difference if you start with P instead of B.)To find the total number of ways, multiply these choices together:  5 x 4 x 3 x 2 x 1  =  120 ways. If there is a difference beween listing them clockwise and anticlockwise, multiply that answer by 2.

Using   R=u^2∗sin(2θ)/g      where R is the range = 20m , u = the intial velocity =  60m/2 , g = 9.8m/s    and θ  = the launch angle  .....we have....

20m = [3600m^2/s^2] sin(2θ) / [9.8m/s^2]

20  = [3600/9.8]sin(2θ)

sin (2θ)  = [20*9.8] / 3600

sin(2θ)  = .54444  = about 1.5605°

Here's the graph of the arrow's path : https://www.desmos.com/calculator/purfk8aeng

lots

Why don't you try it on any calculator, including the one here, and see what you get!!.

x=tan-1 (3.27777778)

Just use any calculator to take the inverse tangent (tan^-1) of 3 5/18=73.03 degrees.

Hallo Gast!

Es wird gesagt, Wilhelm Tell habe einen Apfel vom Kopf seines Sohnes mit einem Pfeil getroffen.  Wenn der Pfeil mit einer Anfangsgeschwindigkeit von 60 m / s abgeschossen wurde und der Junge  20 m entfernt war, mit welchem Abflugwinkel hat Bill zielen müssen? (Es sei angenommen, dass  Pfeil und Apfel zunächst auf gleicher Höhe über dem Boden sind.)

v = 60 m/s

a = 20 m

c = Flugstrecke des Pfeiles

h = Fallstrecke des Pfeiles

t = Flugzeit

α = Abflugwinkel

c = v * t

t = c / v

c = a / cos α

h = (g / 2) * t²

h = a * tan α

a * tan α = (g / 2 ) * t²

t = √(2a * tan α / g)

v * t = a / cos α

t = a / v * cos α

√(2a * tan α / g) = a / v * cos α

2a * tan α / g = a² / v² * cos² α                 cos² α = 1 / (1 + tan² α)

2a * tan α / g = a² * (1 + tan² α ) / v²

40m * tan α / 9,81m/s² = 1600m² * s² * (1 + tan² α) / 3600 m²

tan α = x

4,0775 s² * x = 0,4444 s² * (1 + x²)

0,4444 * x² - 4,0775x + 0,4444 = 0

x² - 9,1743 + 1 = 0

x(1;2) = 4,5872 +- √(21,0420 - 1)

x(1;2) = 4,5872 +- 4,4768

x1 = 0,1104 = tan α1

α1 = 6,3°

x2 = 9,0640 = tan α2

α2 = 83,7°

Bill can shoot under 6,3° or 83,7°

Greetings asinus :- )

!

45,480+767 =46,247. Learn to use the calculator here.

5 x+3>=x-9 subtact x from both sides

4x + 3>= -9 subtract 3 from both sides

4x >= -12 divide both sides by 4

x >= -3

 6y+3 ≤ -15 subtract 3 from both sides

6y  ≤ -18 divide both sides by 6

y ≤ - 3

1.   [9, 12, 16]       the triangle is possible

The triangle   9-12-15  is a right triangle with the angle opposite the longest side = 90°

Therefore.......the angle opposite the longest side here will be > 90°  since 16 >15

Then, this trangle is obtuse  [we can prove this using the Law of Cosines]

2. (11,14, and 27)   this triangle is not possible

Remember that the sum of any two sides of a triangle must be greater than the remaining sde.....but....     11 +  14   = 25         and 25 <  27       ....so......no triangle is formed

3. (1.5,3.6, and 3.9)   

Remember that, if we multiply every side by 10, we will have a similar triangle to the one given....so.....this triangle will be    15 - 36 - 39

Note   15^2  + 36^2  = 39^2   ....so    this triangle is a right triangle

4. (2,3.7, and 4.1)      multiply by 10 again and we have    20 - 37 - 41

Note that   sqrt[20^ + 37^2]   =  about 42.05

So.....since  a right triangle is formed  with sides of 20, 37 and   ≈ 42.05

And the angle opposite the right triangle with a side of about 42.05 would be 90°

But this triangle has a third side which  is < 42.05

Thus.....the angle opposite the side 41 in a 20 - 37 - 41 triangle will be < 90°

So.....this triangle is acute because the other two angles will also be acute [we can prove this using the Law of Sines]

16.50+6.60 +14.30=37.40

Compute the definite integral:

integral_(-1)^1 1/sqrt(1-x^2) dx

1/sqrt(1-x^2) has discontinuities at x = -1 and x = 1, which both produce improper bounds:

= integral_(-1)^1 1/sqrt(1-x^2) dx

Since 1/sqrt(1-x^2) is an even function and the interval [-1, 1] is symmetric about 0, integral_(-1)^1 1/sqrt(1-x^2) dx = 2 integral_0^1 1/sqrt(1-x^2) dx. Applying this identity will reduce the number of improper endpoints on the integration domain:

= 2 integral_0^1 1/sqrt(1-x^2) dx

Apply the fundamental theorem of calculus.

The antiderivative of 1/sqrt(1-x^2) is sin^(-1)(x):

= lim_(b->1^-) 2 sin^(-1)(x)|_0^b

Evaluate the antiderivative at the limits and subtract.

lim_(b->1^-) 2 sin^(-1)(x)|_0^b = (lim_(b->1^-) 2 sin^(-1)(b))-2 sin^(-1)(0) = (lim_(b->1^-) 2 sin^(-1)(b))-0:

= (lim_(b->1^-) 2 sin^(-1)(b))

lim_(b->1^-) 2 sin^(-1)(b) = pi:

Answer: |= pi