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-BARSIC- [3]
4 years ago
5

Numerous engineering and scientific applications require finding solutions to a set of equations. Ex: 8x + 7y = 38 and 3x - 5y =

-1 have a solution x = 3, y = 2. Given integer coefficients of two linear equations with variables x and y, use brute force to find an integer solution for x and y in the range -10 to 10.
Physics
2 answers:
Irina-Kira [14]4 years ago
6 0

Answer:

we do not have the integer coeficients, so the solutions may not be in the range -10 to 10 (it does not happen for all coeficients) so let's do other thing:

Ok, suppose you have the equations:

1) A*x + B*y = C

2) a*x + b*y = c

where A, B, C, a, b and d are knowed.

Let's solve them in the most general way, and you always can use those solutions in the future, just need to replace the values of the constants up there.

First, we isolate x or y in a equation, let's do it with x in equation 1:

A*x  = C - B*y

x = (C - B*y)/A

Now we can replace this in the equation 2, and solve it for y:

a*x + b*y = c

a*(C - B*y)/A + b*y = c

y*(b - a*B/A) = c - a*C/A

y = (c - a*C/A)/(b - a*B/A)

Now you have a solution for y, and we you know the value of y, you can put it in the eqution:

x = (C - B*y)/A

and find the value of x

mote1985 [20]4 years ago
5 0
Need help with this too
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the jamaican bobsled team hit the brakes on their sled so that it decelerates at a uniform rate of 0.43 m/s, how long does it ta
IceJOKER [234]
 <span>here's a cheap trick 
it would take the same time to accelerate from rest to top speed 
as it would take to decelerate from top speed to zero 
so 
instead of 
d = Vi t + 1/2 a t^2 where Vi is positive and a is negative 
we'll use 
Vi = 0 and a is positive 
giving 
85 = 0 + 1/2 (0.43) t^2 = 0.215 t^2 
t^2 = 395.345 
t = 19.88s or 20. s to 2 sig figs 

or we ccould find Vi from 
Vf*2 = Vi^2 + 2 a d 
0 = Vi^2 + 2 (0.43) 85 
Vi^2 = 71.4 
Vi = 8.45m/s 
then 
85 = 8.45 t + 1/2 (-0.43) t^2 
85 = 8.45 t - 0.215 t^2 
0.215 t^2 - 8.45 + 85 = 0 
t = 19.65s or 20. s to 2 s.f.(minor difference arises from rounding Vi) 
or another cheap trick 
when a is constant 
Vavg = (Vf + Vi) /2 = 8.45/2 = 4.225 
and 
d = Vavg t 
85 = 4.225 t 
t = 20.12 or 20. s to 2 s.f. (minor differences from intermidiate roundings) 

anyway you choose you get 20. s</span>
4 0
3 years ago
Define what is boyleâs law?
lawyer [7]

Answer:

Volume of given mass of gas is inversely proportional to pressure of gas

Explanation:

Boyle's law: It states that the volume of given mass of gas is inversely proportional  to the pressure of gas at constant temperature.

Mathematical representation:

Suppose, a gas of mass m

T=Constant temperature

V=Volume of gas

P=Pressure of gas

Then, V\propto\frac{1}{P}

8 0
4 years ago
In the classic horse and cart problem, a horse is attached to a cart that can roll along on a set of wheels. Which of the follow
Marina86 [1]
E is correct because net force in the forward direction is greater
4 0
3 years ago
Three astronauts outside a spaceship and that they decide to play catch. All the astronauts weigh the same on Earth and are equa
guajiro [1.7K]

Answer:

Well concluding there is no gravity their motions would be slow and lightweighted. Let's say they were playing on Earth it would approximately take around 5 to 6 minutes even less, so in space it will approximately take around 10 to 12 minutes may be more but this is just my opinion after using my calculator! Hope this helped!

5 0
3 years ago
Two carts, one twice as heavy as the other, are at rest on a horizontal track. A person pushes each cart for 8 s. Ignoring frict
GalinKa [24]

Answer:

The correct option is: B that is 1/2 K

Explanation:

Given:

Two carts of different masses, same force were applied for same duration of time.

Mass of the lighter cart = m

Mass of the heavier cart = 2m

We have to find the relationship between their kinetic energy:

Let the KE of cart having mass m be "K".

and KE of cart having mass m be "K1".

As it is given regarding Force and time so we have to bring in picture the concept of momentum Δp and find a relation with KE.

Numerical analysis.

⇒ KE =  \frac{mv^2}{2}

⇒ KE =  \frac{mv^2}{2}\times \frac{m}{m}

⇒ KE =  \frac{m^2v^2}{2}\times \frac{1}{m}

⇒ KE =  \frac{(mv)^2}{2}\times \frac{1}{m}

⇒ KE =  \frac{(\triangle p)^2}{2}\times \frac{1}{m}

⇒ KE =  \frac{(\triangle p)^2}{2m}=\frac{(F\times t)^2}{2m}

Now,

Kinetic energies and their ratios in terms of momentum or impulse.

KE (K) of mass m.

⇒ K=\frac{(F\times t)^2}{2m}           ...equation (i)

KE (K1) of mass 2m.

⇒ K_1=\frac{(F\times t)^2}{2\times 2m}

⇒ K_1=\frac{(F\times t)^2}{4m}         ...equation (ii)

Lets divide K1 and K to find the relationship between the two carts's KE.

⇒ \frac{K_1}{K} =\frac{(F\times t)^2}{4m} \times \frac{2m}{(F\times t)^2}

⇒ \frac{K_1}{K} =\frac{2m}{4m}

⇒ \frac{K_1}{K} =\frac{2}{4}

⇒ \frac{K_1}{K} =\frac{1}{2}

⇒ K_1=\frac{K}{2}

⇒ K_1=\frac{1}{2}K

The kinetic energy of the heavy cart after the push compared to the kinetic energy of the light cart is 1/2 K.

7 0
3 years ago
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