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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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6 0
3 years ago
Read 2 more answers
a concave lens creates a virtual image at -47.0 cm and a magnification of +1.75. what is the focal length?
Lilit [14]

The focal length of given concave lens will be -26.85 cm

The height of an image to the height of an object is the ratio that is used to determine a lens' magnification. Additionally, it is provided in terms of object and image distance. It is equivalent to the object distance to image distance ratio.

Given  concave lens creates a virtual image at -47.0 cm and a magnification of +1.75.

We have to find focal length

The focal length can be found out by following way:

Magnification = m = +1.75

m = hi/h

hi = -47 cm

1.75 = -47/h

h = -26.85 cm

So the focal length of given concave lens will be -26.85 cm

Learn more about magnification factor here:

brainly.com/question/6947486

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8 0
1 year ago
MEASUREMENT
lubasha [3.4K]

Answer:

You would need to type the numbers in the question or you could have added a picture but you didn't so there is no way to answer this question. Have a nice day!

5 0
2 years ago
Help with this ASAP it’s due today
adoni [48]
I didn’t know water has calories
7 0
3 years ago
Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a
34kurt

Answer:

E=\frac{KQ}{2\sqrt 2a^2}

Explanation:

We are given that

Charge on ring= Q

Radius of ring=a

We have to find the magnitude of electric filed on the axis at distance a from the ring's center.

We know that the electric field at distance x from the center of ring of radius R is given by

E=\frac{kQx}{(R^2+x^2)^{\frac{3}{2}}}

Substitute x=a and R=a

Then, we get

E=\frac{KQa}{(a^2+a^2)^{\frac{3}{2}}}

E=\frac{KQa}{(2a^2)^{\frac{3}{2}}}

E=\frac{KQa}{2\sqrt 2a^3}

E=\frac{KQ}{2\sqrt 2a^2}

Where K=9\times 10^9 Nm^2/C^2

Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=\frac{KQ}{2\sqrt 2a^2}

4 0
3 years ago
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