Figure 1 is dilated to get Figure 2.

Describe the graph of a system of equations that has no solution.

Thepotemich [5.8K]

The solution for a system of equations is the value or values that are true for all equations in the system. The graphs of equations within a system can tell you how many solutions exist for that system. Look at the images below. Each shows two lines that make up a system of equations.

One SolutionNo SolutionsInfinite SolutionsIf the graphs of the equations intersect, then there is one solution that is true for both equations. If the graphs of the equations do not intersect (for example, if they are parallel), then there are no solutions that are true for both equations.If the graphs of the equations are the same, then there are an infinite number of solutions that are true for both equations.

When the lines intersect, the point of intersection is the only point that the two graphs have in common. So the coordinates of that point are the solution for the two variables used in the equations. When the lines are parallel, there are no solutions, and sometimes the two equations will graph as the same line, in which case we have an infinite number of solutions.

Some special terms are sometimes used to describe these kinds of systems.

The following terms refer to how many solutions the system has.

5 0

3 years ago

Write the standard form of the equation for the circle that passes through the points (2,31),(-15,14),(33,0)

stepladder [879]

Answer:

Step-by-step explanation:

Begin with the standard form of a circle as a conic:

$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$

For a circle, A and C will be the exact same, and B will equal 0. If B is non-zero, the equation represents a rotation of a conic, which is reserved for college-level courses. Shortening this, then:

$x^2+y^2+Dx+Ey+F=0$ is good enough for us for this. Start with the first point on the circle, (2, 31) and fill in the equation above with x and y:

$2^2+31^2+2D+31E+F=0$ which simplifies down to:

$(1):2D+31E+F=-965$

Do the same with the next point on the circle, (-15, 14):

$-15^2+14^2-15D+14E+F=0$ which simplifies down to:

$(2):-15D+14E+F=0$

Do the same with the last point, (33, 0):

$33^2+0^2+33D+0E+F=0$ which simplifies down to:

$(3):33D+F=-1089$

Now we will add (1) and (2) to get (4):

2D + 31E + F = -965

-15D + 14E + F = -421

Multiply the top equatio by -1 to get rid of the F terms:

-2D - 31E - F = 965

-15D + 14E + F = -421

which simplifies to

(4): -17D - 17E = 544

Now add (2) and (3) to get (5):

-15D + 14E + F = -421

33D + F = -1089

Multiply the bottom equation by -1 to get rid of the F terms:

-15D + 14E + F = -421

-33D - F = 1089

which simplifies to

(5): -48D + 14E = 668

Now add (4) and (5) together and eliminate the E terms:

-17D - 17E = 544

-48D + 14E = 668

In order to eliminate the E terms, multiply the top equation by 14 and the bottom equation by 17 to solve for D:

-238D - 238E = 7616

-816D + 238E = 11356

Which gives you that

D = -18

Now plug the value for D into (4) to find E:

-17(-18) - 17E = 544 and

306 - 17E = 544 and

-17E = 238 so

E = -14

Now plug the values for both D and E into (1) to find F:

2(-18) + 31(-14) + F = -965 and

-36 - 434 + F = -965 and

-470 + F = -965 so

F = -495

Now we can fill in the standard form of the conic:

$x^2+y^2-18x-14y=495$

but we're not done til we complete the square on both the x terms and the y terms (and I am assuming you know how to complete the square):

$(x^2-18x+81)+(y^2-14y+49)=495+81+49$ which simplifies to

$(x-9)^2+(y-7)^2=625$

The second choice down matches that equation, but the center they have there is not correct. The center of that circle is (9, 7) and they have it as being (7, 10) with a radius of 5. The radius is 25. The center is wrong in the equation that represents the circle as is the radius. Maybe let someone know that...

4 0

3 years ago

Read 2 more answers

140 divided by 25 equals what? I need an answer fast

Harlamova29_29 [7]

It equals 5.6. ........

8 0

3 years ago

Read 2 more answers

There are 12 people in a class. 10 are randomly chosen. How many possible combinations are there?

Mazyrski [523]

long way