The diffrence between 2 numbers is 1 and their product Is 132

Mathematics

1 answer:

il63 [147K]3 years ago

6 0

Answer:

x = 12

y = 11

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Step-by-step explanation:

$Let -the-unkonown- numbers-be-x-and-y \\x-y = 1 (1)\\xy = 132 (2)\\Make-x-subject-of-formula-in-equation1 \\x = 1+y\\Substitute , 1+y for- x -in -equation ,2\\y(1+y) = 132\\y + y^{2} =132\\y^{2} +y-132=0\\Solve- the-quadratic-equation.\\y = 11\\substitute- 11 -for- y- in- equation -1- 0r- 2\\x -y = 1\\x-11 = 1\\x = 1+11\\x = 12$

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A 1 newton force will stretch a spring 1 meter. The spring/mass system is damped by a force that is 8 times the instantaneous ve

sukhopar [10]

Answer:

Step-by-step explanation:

Given the mass is m =16kg, and 1N force will stretch the spring 1 m.

That is, F =1N,Z =1m. Now find the spring constant k:

F = kL = 1 = k(1) = k= 1N/m.

The damping force is 8times the instantaneous velocity, this means β = 8,

and the external force is f(t) = 0

Initially the object compressed 0.6m above equilibrium position,

with the downward velocity is 2m/s.

The differential equation for a spring mass system with

damping force and extemal force is: mx" + βxt + kx = f(t).

so, 16x"+ 8x' + x= 0, x(0} = -0.6, x'(0)= 2m/s.

Now solve the DE:

The auxilary equation for the homogeneous equation is 16x"+8x'+x=0

solving we get, 16r² + 8r + 1 = 0 => (4r + 1)² = 0 => r = - 1/4.

Then the general solution for the homogenous system is: $x(t)=c_1e^{-t/4} +c_2te^{-t\4}$ .

Use the initial conditions x (0) = -0.6, x'(0) = 2m/s:

$x(0)=c_1e^{0} +c_2(0)e^{0}=-0.6=c_1\\x'(t)=-\frac{1}{4}c_1e^{-t/4}+c_2e^{-t/4}-\frac{1}{4}c_2te^{-t/4}\\x'(0)=-\frac{1}{4}c_1e^0+c_2e^0-\frac{1}{4}c_2(0)e^0=2=-\frac{1}{4}(-0.6)+c_2=c_2=1.85$ .