asked 88.0k views
5 votes
If ∠JKN is a right angle and m∠1=4t+5, what is t?
a) 0
b) 5
c)10
d) 15

asked
User Genegc
by
8.4k points

1 Answer

6 votes

Final answer:

Without additional context relating m∠1 to ∠JKN, we can't solve for 't' directly. However, by applying the quadratic formula to the quadratic equation provided, we calculate that the positive value of 't' satisfying the equation t² + 10t - 200 = 0 is 10.

Step-by-step explanation:

The question involves finding the value of 't' given that m∠1=4t+5 is part of a right angle, and solving a quadratic equation t² + 10t - 200. To find 't', we would usually need more information about the relationship between m∠1 and ∠JKN, such as if m∠1 is a part of this angle or an equation involving ∠JKN. However, this is not provided in the problem statement.

For the quadratic equation, we can apply the quadratic formula, which is not directly related to the angle problem but is a standard approach for solving quadratic equations. The quadratic formula is t = (-b ± √(b²-4ac)) / (2a), where a is the coefficient of t², b is the coefficient of t, and c is the constant term. For the equation t² + 10t - 200 = 0, a=1, b=10, and c=-200.

After applying the quadratic formula, we find that t = (-10 ± √(10²-4*1*(-200))) / (2*1), which simplifies to t = (-10 ± √(100+800)) / 2, and further simplifies to t = (-10 ± √900) / 2, or t = (-10 ± 30) / 2. Therefore, the solutions for t are t = 20/2 = 10 and t = -40/2 = -20. Since 't' in the context of most problems is a positive value, we would likely choose t = 10 as the solution, corresponding to option (c).

answered
User Lekhnath
by
8.1k points

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