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Vlad1618 [11]
3 years ago
12

Heather writes the equations below to represent two lines drawn on the coordinate plane. –6x + 18y = 0 4x – 12y = 20 After a

pplying the linear combination method, Heather arrived at the equation 0 = 60. What conclusion can be drawn about the system of equations?

Physics
2 answers:
devlian [24]3 years ago
8 0

Answer:

The system has no solution

Explanation:

You can check it graphically, the pair of equations represents two lines, if they never cross, or if they are parallel, the system of equations is inconsistent and has no solution.

I attached you two pictures in which you can see the graphical result.

yuradex [85]3 years ago
5 0

Answer:

The equation has no solution; therefore, the system of equations has no solution.

Explanation:

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zheka24 [161]

Answer: A if thats not right its C

Explanation:

3 0
3 years ago
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An object that is falling has the following type(s) of energy. Ignore air resistance.
Anton [14]
Potential and kinetic
6 0
3 years ago
Read 2 more answers
You’ve made the hypothesis that the stepper the slope , the faster a ball will be rolling when it reaches the bottom .
BigorU [14]

Answer:

B) how steep the slope is

Explanation:

Because you have to know how is the influence of the steep of the slope in the time that a ball reaches the bottom. The steep of the slope is the variable that you would have to change in an experiment.

I hope this is useful for you

regards

4 0
3 years ago
One factor that could account for a drop in the water table is:
Shtirlitz [24]

Answer:

D. Less rain and snow.  

Explanation:

A factor that can be a account for a drop in the water table is, less rain and snow. to topography, water tables is influenced by lot of factors, including the geology, weather, ground cover.

4 0
3 years ago
A rocket car on a horizontal rail has an initial mass of 2500 kg and an additional fuel mass of 1000 kg. At time t0 the rocket m
slamgirl [31]

Answer: Acceleration of the car at time = 10 sec is 108 m/s^{2} and velocity of the car at time t = 10 sec is 918.34 m/s.

Explanation:

The expression used will be as follows.

M\frac{dv}{dt} = u\frac{dM}{dt}

\int_{t_{o}}^{t_{f}} \frac{dv}{dt} dt = u\int_{t_{o}}^{t_{f}} \frac{1}{M} \frac{dM}{dt} dt

       = u\int_{M_{o}}^{M_{f}} \frac{dM}{M}

v_{f} - v_{o} = u ln \frac{M_{f}}{M_{o}}

v_{o} = 0

As, v_{f} = u ln (\frac{M_{f}}{M_{o}})

u = -2900 m/s

M_{f} = M_{o} - m \times t_{f}

           = 2500 kg + 1000 kg - 95 kg \times t_{f}s

           = (3500 - 95t_{f})s

v_{f} = -2900 ln(\frac{3500 - 95 t_{f}}{3500}) m/s

Also, we know that

     a = \frac{dv_{f}}{dt_{f}} = \frac{u}{M} \frac{dM}{dt}

        = \frac{u}{3500 - 95 t} \times (-95) m/s^{2}

        = \frac{95 \times 2900}{3500 - 95t} m/s^{2}

At t = 10 sec,

v_{f} = 918.34 m/s

and,   a = 108 m/s^{2}

3 0
3 years ago
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