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y = (5/7) (3-1) + 1 = 2.428571429

Don't be lazy, what is ((3m-9)(m^2+5m+6))/((4m+8)(m^2-9))

if you solve well it is 3/4

Beeinflusst die Schwerebeschleunigung auf der Erde und auf dem Mond die periodische Schwingung des einfachen Pendels? Beweisen Sie Ihre Antwort mit einer Gleichung.

Does the gravitational acceleration on earth and on the moon influence the periodic oscillation of the simple pendulum? Prove your answer with an equation.

Harmonische Schwingung eines mathematischen Pendels

bei kleinen Amplituden.

Harmonic oscillation of a mathematical pendulum
with small amplitudes.

Die Periodendauer errechnet sich mit

The period is calculated using

\(t_0=2\pi\sqrt{\frac{l}{g}}\)

Die Fallbeschleunigung ist

The acceleration due to gravity is

auf der Erde        \(g_{(erde)}=9.81m/s^2\)

auf dem Mond    \(g_{(mond)}=1.6m/s^2\)

Die Periodendauer ist bei einer Pendellänge von einem Meter

The period is with a pendulum length of one meter

auf der Erde   \(t_{0(erde)}=2\pi \sqrt{\frac{1m\cdot s^2}{9.81m}}\\ \color{blue}t_{0(erde)}=2\ Sekunden \)    on earth

auf dem Mond    \(t_{0(mond)}=2\pi \sqrt{\frac{1m\cdot s^2}{1.6m}}\\ \color{blue}t_{0(mond)}=5\ Sekunden \)  on the moon

Die verschiedenen Größen der Schwerebeschleunigung auf Erde und Mond beeinflussen die Periodendauer eines Pendels.

The different magnitudes of the gravitational acceleration on earth and moon influence the period of a pendulum.

Übersetzungshilfe:

What is an equation of the line that passes through the point (-4,5) and is perpendicular to the line 4x+y=7?

Hello Guest!

\(4x+y=7 \)

\(y=-4x+7\)

\(m=-\frac{1}{(-4)}\)

\(m=\frac{1}{4}\)

\(y=m(x-x_1)+y_1\)

\(y=\frac{1}{4}\cdot (x-(-4))+5\\ y=\frac{1}{4}x+1+5\)

\(y=\frac{1}{4}x+6\)

  !

We are given that the solids are made from the same material, which means they have the same density.

Since   mass   =   density * volume

mass_A / mass_B   =   (density_A * volume_A) / (density_B * volume_B)

                                                                                                           density_A = density_B   so we can cancel them out

mass_A / mass_B   =   volume_A / volume_B

So here we have shown that the ratio of the masses is equal to the ratio of their volumes.

Now we need to be careful and remember that the ratio of their volumes is not equal to the ratio of their surface areas.

The ratio of their volumes is equal to the ratio of their surface areas raised to the power of 3/2

Find an equation in the form y = ax^2 +bx+c for the parabola passing through the points.(-3,57), (1,1), (-2,31)

Hello Guest!

With three points one can determine a quadratic function of the form f (x) = ax2 + bx + c. You insert the points and then solve the linear system of equations.

  !

(x + 1)^3 > (x + 1)^2.                 Why not START by dividing both sides by  (x+1)²     ??   Try that.....

All of the angles inside the triangle sum to 180^o

180 - 50 = 130      this is what the two unknown anglse sum to

2x  +    3x-10   = 130^o          now you should be able to easily solve for 'x'

Piecewise:

   1 + 3/4 =  7/4

1 / (7/4)   = 4/7

1 + 4/7 = 11/7

1 / (11/7)  = 7/11

minus 51

Let WXYZ be a trapezoid with bases XY and WZ. In this trapezoid ANGLE ZXW=105 DEGREES, angle XWZ=43 degrees, and angle XYZ= 141 degrees. Find the angle YXZ, in degrees.

This is a hint. 

Thank you so so much! I missed my class and didn't know what to do.

First....realize this is a right triangle

   Pythagorean theorem

       (sqrt5)^2 + (sqrt8)^2 = sqrt13 ^2  

area = 1/2 b * h    = 1/2 (sqrt5 * sqrt8 ) = sqrt 10  units²

Angle XYZ = 83 degrees.

Using long division, you get 0.(b - 1) (b - 1) (b - 1) ...., where b - 1 is an actual digit in base b.

1000 / 4 =250 mph - its speed with a headwind.
1000 / 3 1/13 =300 mph - its speed with the wind.

[250 + 300] / 2 =275 mph - its speed in still air.
[300 - 250] / 2  =25 mph - wind's speed

x+y = 124

5x + 10y = 840        x = # 5 dollar bills     y = #10 dollar bills

    solve for x and y

0.72*31 + 0.09N =0.36*[31 + N], solve for N

N =124 / 3 = 41+1/3 oz. of 9% acid will be needed.

a + b =10, 

a * b = -6, solve for a, b

a = 5 - sqrt(31)  AND  b = 5 + sqrt(31)

a = 5 + sqrt(31)  AND b = 5 - sqrt(31)

As far as I know, the only way to check thousands of numbers to see if their digits sum up to 20 is to to write a short computer code and check them one by one. I wrote just such a programme and counted ALL the numbers that sum up to 20, and the 1000th number is =15,770.

The smallest integer n = 17

17^(-1) mod 130 = 23

17^(-1) mod 231 = 68

Using CRT and MMI:

x mod 6 = 4
x mod 9 = 4
x mod 15 = 4
x mod 18 = 4, solve for x

x =[90 x 4] + 4= 364 - the smallest multiple of 7 that satisfies the 4 congruences.

December has 31 days - November has 30 days = 1 day x 24 hours x 60 minutes x 60 seconds =86,400 seconds - the remainder when you divide number of seconds in December by number of seconds in November.

86,400 / 11 =7,854 this is the quotient.

86,400 - [7,854 x 11] = 6 this is the remainder.

Or, you can do it directly like this:

86,400 mod 11 = 6 - the remainder.

wrong question