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11 days 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 [913]11 days 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 [1K]11 days ago

5 0

I also require assistance with this.

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Pilobolus is a genus of fungi commonly found on dung and known for launching its spores a large distance for a sporangiophore on

Softa [913]

1) The work performed on the spores is $2.45\cdot 10^{-7}J$

2) The spores land 0.32 m away

Explanation:

1)

According to the work-energy principle, the work done on the spores equals their change in kinetic energy:

$W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where

W is work done

m represents the spore's mass

v is the final velocity

u is the initial velocity

Given:

spore mass = $m=10^{-8} kg$

initial velocity u = 0 (starting from rest)

final velocity v = 7.0 m/s

Calculating the work:

$W=\frac{1}{2}(10^{-8})(7.0)^2-0=2.45\cdot 10^{-7}J$

2)

The spores follow projectile motion, moving along a parabolic trajectory made of two motions:

- Constant speed horizontally

- Vertically accelerated due to gravity

Considering vertical motion first, using kinematics:

$s=ut+\frac{1}{2}at^2$

where

s = 0.01 m (shooting height)

u = 0 (no initial vertical velocity, horizontal ejection)

t = time of flight

g = acceleration due to gravity $a=g=10 m/s^2$

Solving for t:

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(0.01)}{10}}=0.045 s$

Now for horizontal displacement, when velocity is constant:

$d=v_x t$

where

horizontal velocity = $v_x = 7.0 m/s$

time t = 0.045 s

Calculating distance d:

$d=(7.0)(0.045)=0.32 m$

Thus, the spores land 0.32 meters away.

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

3 0

13 days ago

A box weighing 46 newtons rests on an incline that makes an angle of 25° with the horizontal. What is the magnitude of the compo

inna [987]

The force of the box’s weight acting perpendicular to the slope can be computed using the formula:

F = wcos(a)

In this equation, F represents the component of the weight perpendicular to the slope,

W denotes the box's total weight,

and A is the angle of the slope.

Thus, substituting values gives: F = (46)cos(25)

Resulting in F = 42 N

8 0

3 days ago

A hydraulic lift raises a 2000 kg automobile when a 500 N force is applied to the smaller piston. If the smaller piston has an a

kicyunya [1025]

Answer:

The cross-sectional area of the larger piston is 392cm ^{2}[/tex]

Explanation:

To find the solution, we apply the following equation:

Pascal's principle: F=P*A Formula (1)

F=Force applied to the piston

P: Pressure

A= Area of the piston

Nomenclature:

Fp= Force on the primary piston= 500N

W= weight of the car =m*g=2000kg*9.8m/s2= 19600N

Fs= Force on the secondary piston= W = 19600N

$Ap= Primary piston area=10cm^{2} =10*10^{-4}m^{2}$

As= Area of the secondary piston=?

Pressure applied on one side is distributed to all liquid molecules since liquids are incompressible.

From equation (1)

P=F/A

Pp=Ps

$\frac{Fp}{Ap} = \frac{Fs}{As}$

$As= \frac{Fs*Ap}{Fp}$

$As=\frac{19600*10*10^{-4} }{500}$

$As=0.0392m^{2} =0.0392*10^{4}cm^{2}$

$As=392cm ^{2}$

5 0

9 days ago

Suppose you push a hockey puck of mass m across frictionless ice for a time 1.0 s, starting from rest, giving the puck speed v a

Softa [913]

According to Newton's second law, Force equals the rate of change of momentum over time. Momentum change is equal to Force times time. So, F=ma can be rearranged to a=F/m, a more recognizable formulation of Newton's second law
Using a relevant kinematic equation for mass m: V=u+at; where initial speed u=0; thus, acceleration a=F/m gives V=(F/m)xt, which translates to t=mV/F. For mass 2m, applying the same formula: V=u+at; u=0; a=F/2m indicates V=(F/2m)xt, leading to t=2mV/F (possibly double the initial time)
I might have erred somewhere along the line, but the fundamental concept seems valid... using another kinematic equation for m: s=ut + (1/2)at²; with s=d; and initial speed u=0; a=F/m; t=1; results in d=(1/2)(F/m) = F/2m. Similarly, for 2m: s=ut + (1/2)at²; s=d; u=0; a=F/2m; and t=1 gives d=(1/2)(F/2m)=F/4m (half the distance perhaps???? WHAT???!)

3 0

11 days ago

Read 2 more answers

Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimet

ValentinkaMS [1149]

Answer:

The rate at which the root beer level is decreasing is 0.08603 cm/s.

Explanation:

The formula for the volume of the cone is:

$V=\frac {1}{3}\times \pi\times r^2\times h$

Where V denotes the cone's volume

r indicates the radius

h signifies the height

The ratio of radius to height remains consistent throughout the cone.

Thus, we have r = d / 2 = 10 / 2 cm = 5 cm

h is 13 cm

Consequently, r / h = 5 / 13

r = {5 / 13} h

$V=\frac {1}{3}\times \frac {22}{7}\times ({{{\frac {5}{13}\times h}}})^2\times h$

$V=\frac {550}{3549}\times h^3$

Additionally, we differentiate the volume expression in relation to time:

$\frac {dV}{dt}=\frac {550}{3549}\times 3\times h^2\times \frac {dh}{dt}$

Given that $\frac {dV}{dt}$ = -4 cm³/sec (the negative sign indicates outflow)

h equals 10 cm

Hence,

$-4=\frac{550}{3549}\times 3\times {10}^2\times \frac {dh}{dt}$

$\frac{55000}{1183}\times \frac {dh}{dt}=-4$

$\frac {dh}{dt}=-0.08603\ cm/s$

The rate at which the root beer level is decreasing is 0.08603 cm/s.

3 0

11 days ago