do u have to use the exact same number for all the substitutions?
The solutions to the given quadratic equation {49n² - 301n + 42 = 0 } are 1/7 and 6.
<h3>What are the solutions to the given quadratic equation?</h3>
Quadratic equation is simply an algebraic expression of the second degree in x. Quadratic equation in its standard form is expressed as;
ax² + bx + c = 0
Where x is the unknown
To solve for x, we use the quadratic formula
x = (-b±√(b² - 4ac)) / (2a)
Given the equation in the question;
49n² - 301n + 42 = 0
Compared to the standard form of quadratic equation { ax² + bx + c = 0 }
We plug in these values into the quadratic formula.
x = (-b±√(b² - 4ac)) / (2a)
x = (-(-301) ±√((-301)² - 4 × 49 × 42 )) / (2 × 49)
x = ( 301 ±√( 90601 - 8232 )) / 98
x = ( 301 ±√( 82369 )) / 98
x = ( 301 ± 287) / 98
x = (301 - 287)/98, (301 + 287)/98
x = 14/98, 588/98
x = 1/7, 6
Therefore, the solutions to the given quadratic equation {49n² - 301n + 42 = 0 } are 1/7 and 6.
Learn more about quadratic equations here: brainly.com/question/1863222
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Answer:
The second graph.
Step-by-step explanation:
A graph with a relation that is not a function would have more than one y value at any x point. For example, in the second graph, at (-0.5) on the x-axis, the y values are 4 and -4 so its relation does not have a function.
The answer is C.
This is because 7 must be greater than 12-g.
G cannot equal 5 and below because 12-5=7.
Therefore C is the only answer that works.
Don’t look up that link they put the same thing on another question
