Which graph shows a system with an infinite number of solutions?
1 answer:
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Answer:
Step-by-step explanation:
The zeros of a function is where the function crosses the x-axis. The x-intercepts are the zeros of a function.
Let output equal to 0.
Set factors equal to 0.
The zeros of the function are and
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1. Assuming OP || NM, considering the triangle proportionality theorem possible measurements for the segments are: ON = 3 units, LO = 1 unit, PM = 6 units, LP = 2 units
2. The possible values are decided by assuming using the triangle proportionality theorem.
3. Another method is applying the AA Similarity Theorem to compare corresponding side lengths of the triangles that are proportional to each other.
<h3>What is the Triangle Proportionality Theorem?</h3>Triangle proportionality theorem states that if a line is parallel to one of the sides of a triangle intersects two of the other two sides, it divides the two sides into proportional corresponding segments.
1. Assuming that OP || NM, we would apply the triangle proportionality theorem to have the following ratio:
Using the above ratio, we can choose the following possible lengths for the segments:
ON = 3 units
LO = 1 unit
PM = 6 units
LP = 2 units
2. The values are decided assuming that the ratio,
This means that the ratio of the proportional segments, irrespective of the their measurement should give you an assumed equal ratio of 3:1.
Therefore, since:
The measurements, ON = 3 units, LO = 1 unit, PM = 6 units, LP = 2 units are therefore possible measurements.
3. Using the AA Similarity Theorem to prove that ΔLPO ≅ ΔLMN, the corresponding sides of both triangles will be proportional.
LP/LM = LO/LN = PO/MN
We can use this to choose possible measurements while assuming a ratio between the proportional sides of both triangles.
Let's assume that: LP/LM = LO/LN = PO/MN = 1/4
Let,
LO = 1 unit
LN = 4 units
Therefore, ON = LN - LO = 4 - 1 = 3 units.
Same way the other possible measures can be gotten provided a ratio is assumed.
Learn more about triangle proportionality theorem on:
