The equation 3x - 2y = 8 and 6x - 4y = 16 are systems of equations with infinite solutions.
A linear system of equations can have (a)unique solutions, (b) infinite number of solutions, or (c) no solution.
For it to have infinite number of solutions, the graph of the equations must intercept with each other at infinite points, leaving the two graphs overlaying each other.
Given the equation, 3x - 2y = 8, the other equation in a linear system to have infinite number of solutions with it must be the exact same line.
To do this, simply multiply the whole equation with a real number.
Example:
Multiplying the whole equation by 2 gives 6x - 4y = 16.
Now, 3x - 2y = 8 and 6x - 4y = 16 are equations in a linear system with infinite number of solutions.
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Answer:
y = x^2/ 60 + 15
=>( x - h)^2 = 4a[ (x^2/6 + 15) - k ].
Step-by-step explanation:
Okay, in order to solve this question very well, one thing we must keep at the back of our mind is that the representation for the equation of a parabola is given as ; y = ax^2 + bx + c.
That is to say; y = ax^2 + bx + c is the equation for a parabola. So, we should be expecting our answer to be in this form.
So, from the question above we are given that "the satellite dish will be in the shape of a parabola and will be positioned above the ground such that its focus is 30 ft above the ground"
We will make an assumption that the point on the ground is (0,0) and the focus is (0,30). Thus, the vertex (h,k) = (0,15).
The equation that best describes the equation of the satellite is given as;
(x - h)^2 = 4a( y - k). ------------------------(1).
[Note that if (h,k) = (0,15), then, a = 15].
Hence, (x - 0)^2 = (4 × 15) (y - 15).
x^2 = 60(y - 15).
x^2 = 60y - 900.
60y = x^2 + 900.
y = x^2/ 60 + 15.
Hence, we will have;
(x - h)^2 = 4a[ (x^2/6 + 15) - k ].
Answer:
4
Step-by-step explanation:
3- 3
23 - 3
20 / 5 = 4.
4 x 5 + 3 = 23, so X is 4.
Answer:
<em>N </em>is located in the third quadrant
Step-by-step explanation:
since the y coordinate is a negative and so is the x coordinate, the negative y would push <em>N </em>down and the negative x would push it left, landing the coordinates of <em>N</em> in the 3rd quadrant.
