Ask question

Ask question

All categories

2 months 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.

2 answers:

Softa [3K]2 months ago

6 0

Answer:

Since we lack the integer coefficients, the solutions could be outside the -10 to 10 range (this isn't the case for every coefficient). Let's proceed differently:

Assume you have the following equations:

1) A*x + B*y = C

2) a*x + b*y = c

where A, B, C, a, b, and c are known.

We will solve this in a general manner, and you can apply these solutions in the future by simply substituting the values for the constants above.

To start, we can express x in terms of y using equation 1:

A*x = C - B*y

x = (C - B*y) / A

Now, we can substitute this expression for x into equation 2 and solve for y:

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)

With this expression for y, you can then find x by substituting y back into the equation:

x = (C - B*y) / A

and determine the value of x.

Maru [3.3K]2 months ago

5 0

I also require assistance with this.

You might be interested in

A pile driver drops from a height of 35 meters

Keith_Richards [3271]

It is all right, so what is it that they said didn't. Fiend. Diehard d fish

3 0

16 days ago

Read 2 more answers

2. An airplane traveling north at 220. meters per second encounters a 50.0-meters-per-second crosswind

Yuliya22 [3333]

The speed resulting from the plane is (3) 226 m/s

Reasoning:

We can determine the plane's resultant speed using the Pythagorean Theorem because the two speeds form a right angle (creating a right triangle).

Thus, the computation is as follows:

$ResultantSpeed=\sqrt{VerticalSpeed^{2}+HorizontalSpeed^{2}}\\\\ResultantSpeed=\sqrt{(220\frac{m}{s})^{2}+50\frac{m}{s})^{2}$

$ResulntantSpeed=\sqrt{48400\frac{m^{2} }{s^{2} }+2500\frac{m^{2} }{s^{2} } } \\\\ResultantSpeed=\sqrt{50900\frac{m^{2} }{s^{2} }}=226\frac{m}{s}$

Consequently, the plane's resultant speed is (3) 226 m/s

Have a wonderful day!

5 0

1 month ago

At time t=0 a proton is a distance of 0.360 m from a very large insulating sheet of charge and is moving parallel to the sheet w

serg [3582]

Explanation:

The formula for the electric field produced by an infinite sheet of charge is outlined below.

E = $\frac{\sigma}{2 \epsilon_{o}}$

where, $\sigma$ is the surface charge density

Following this, the formula for the electric force acting on a proton is given as:

F = eE

where, e is the charge of a proton

According to Newton's second law of motion, the overall force on the proton can be expressed as follows.

F = ma

a = $\frac{eE}{m}$

= $\frac{e(\frac{\sigma}{2 \epsilon_{o}})}{m}$

= $\frac{e \sigma}{2m \epsilon_{o}}$

According to kinematic equations, the proton's speed in the perpendicular direction can be described as follows.

$v_{f} = v_{i} + at$

= $(0 m/s) + \frac{e \sigma}{2 m \epsilon_{o}}t$

= $\frac{1.6 \times 10^{-19}C \times 2.34 \times 10^{-9} C/m^{2} \times 5.40 \times 10^{-8}s}{2 \times (1.67 \times 10^{-27} kg)(8.85 \times 10^{-12} C^{2}/Nm^{2}}$

= 683.974 m/s

Thus, the overall speed of the proton can be calculated as follows.

v' = $\sqrt{(960 m/s)^{2} + (683.974 m/s)^{2}}$

= $\sqrt{921600 + 467820.43}$

= $\sqrt{1389420.43}$

= 1178.73 m/s

Consequently, we conclude that the proton's speed is 1178.73 m/s.

3 0

20 days ago

A shuttle on Earth has a mass of 4.5 E 5 kg. Compare its weight on Earth to its weight while in orbit at a height of 6.3 E 5 met

ValentinkaMS [3465]

Response:

83%

Clarification:

At the surface, the weight can be expressed as:

W = GMm / R²

where G denotes the gravitational constant, M represents the Earth's mass, m signifies the shuttle's mass, and R is the Earth's radius.

When in orbit, the weight is given by:

w = GMm / (R+h)²

where h indicates the shuttle's altitude above Earth's surface.

The weight ratio is as follows:

w/W = R² / (R+h)²

w/W = (R / (R+h))²

For R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

Thus, the shuttle maintains 83% of its weight as it orbits.

4 0

1 month ago

A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then

serg [3582]

Response:

A) $W_{ff} =-744.12J$

B) $F_f=-W_{ff}*sin\theta /hy = 112.75N$

C) $F_{f2}=207.58N$

Clarification:

The question is not fully provided. The complete question was:

A skateboarder with a mass of ms = 54 kg is at the top of a ramp with a height of hy = 3.3 m. He then jumps on his skateboard and goes down the ramp. His speed at the base is vf = 6.2 m/s.

Part (a) Formulate an expression for the work, Wf, done by the frictional force on the skateboarder in terms of the variables listed in the problem.

Part (b) The ramp is at an angle θ with the ground, where θ = 30°. Formulate an expression for the frictional force's magnitude, fr, between the skateboarder and the ramp.

Part (c) Upon reaching the bottom, the skateboarder continues with speed vf onto a grass-covered flat surface. The friction between the grass and the skateboarder brings him to a halt after 5.00 m. Determine the frictional force, Fgrass in newtons, between the skateboarder and the grass.

For part A), we assess the energy balance to determine the work done by the friction:

$W_{ff}=\Delta E$

$W_{ff}=1/2*m*vf^2-m*g*hy$

$W_{ff}=-744.12J$

For part B), we utilize the previously calculated work:

$W_{ff}=-F_f*(hy/sin\theta)$ Rearranging for friction force:

$F_f=-W_{ff}*sin\theta /hy$

$F_f=112.75N$

For part C), we first ascertain the acceleration through kinematic equations and subsequently find the frictional force using dynamic methods:

$Vf^2=Vo^2+2*a*d$

Rearranging for 'a':

$a=-3.844m/s^2$

Now, using dynamics:

$|F_f|=|m*a|$

$|F_f|=207.58N$

8 0

1 month ago