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-BARSIC- [3]
3 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]3 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]3 years ago
5 0
Need help with this too
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The best and most correct answer among the choices provided by your question is the third choice or number 3.

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2 years ago
Rick shoots a basketball at an angle of 35' from the horizontal. It leaves his hands 7 feet from the ground with a velocity of 2
Korvikt [17]

Given:

The angle of projection of the basketball, θ=35°

The height at which the ball leaves the hand, h=7 ft

The initial velocity of the basketball, v=20 ft/s

To find:

The parametric equations describing the shot.

Explanation:

The range, x of the basketball is given by,

x=v\cos\theta t

On substituting the known values,

\begin{gathered} x=20\times\cos35\degree\times t \\ \implies x=16.4t \end{gathered}

The change in the height, y of the basketball is given by,

y=-v\sin\theta t+\frac{1}{2}gt^2

Where g is the acceleration due to gravity.

On substituting the known values,

\begin{gathered} y=-20\times\sin35\degree\times t+\frac{1}{2}\times32\times t^2 \\ \implies y=-11.5t+16t^2 \end{gathered}

Final answer:

The parametric equations describing the shot are

\begin{gathered} \begin{equation*} x=16.4t \end{equation*} \\ \begin{equation*} y=-11.5t+16t^2 \end{equation*} \end{gathered}

8 0
1 year ago
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