The constant should be added to form a perfect square trinomial will be 1/4. Then the correct option is D.
<h3>What is a quadratic equation?</h3>
It's a polynomial with a value of zero. There exist polynomials of variable power 2, 1, and 0 terms. A quadratic equation is an equation with one statement in which the degree of the parameter is a maximum of 2.
The expression is x² + x.
Then the constant should be added to form a perfect square trinomial.
Then the constant will be
The square of the half of the coefficient of the variable x is to be added to make a perfect square.
Then the constant will be 1/4.
Then the perfect square will be
More about the quadratic equation link is given below.
brainly.com/question/2263981
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5 mph to get that answer you use the speed equation
speed(mph)=distance(in miles)/time(in hours)
2.5
------ = 5 mph
.5
Answer:
Tn = 2Tn-1 - Tn-2
Step-by-step explanation:
Before we can generate the recursive sequence, we need to find the nth term of the given sequence.
nth term of an AP is given as:
Tn = a+(n-1)d
If a17 = -40
T17 = a+(17-1)d = -40
a+16d = -40 ...(1)
If a28 = -73
T28 = a+(28-1)d = -73
a+27d = -73 ...(2)
Solving both equations simultaneously using elimination method.
Subtracting 1 from 2 we have:
27d - 16d = -73-(-40)
11d = -73+40
11d = -33
d = -3
Substituting d = -3 into 1
a+16(-3) = -40
a - 48 = -40
a = -40+48
a = 8
Given a = 8, d = -3, the nth term of the sequence will be
Tn = 8+(n-1) (-3)
Tn = 8+(-3n+3)
Tn = 8-3n+3
Tn = 11-3n
Given Tn = 11-3n and d = -3
Tn-1 = Tn - d... (3)
Tn-1 = 11-3n +3
Tn-1 = 14-3n
Tn-2 = Tn-2d...(4)
Tn-2 = 11-3n-2(-3)
Tn-2 = 11-3n+6
Tn-2 = 17-3n
From 3, d = Tn - Tn-1
From 4, d = (Tn - Tn-2)/2
Equating both common difference
(Tn - Tn-2)/2 = Tn - Tn-1
Tn - Tn-2 = 2(Tn - Tn-1)
Tn - Tn-2 = 2Tn-2Tn-1
2Tn-Tn = 2Tn-1 - Tn-2
Tn = 2Tn-1 - Tn-2
The recursive formula will be
Tn = 2Tn-1 - Tn-2
Answer:
It's gonna be 2t - 7 my friend
Step-by-step explanation:
I solved this using a scientific calculator and in radians mode since the given x's is between 0 to 2π. After substitution, the correct pairs
are:
cos(x)tan(x) – ½ = 0
→ π/6 and 5π/6
cos(π/6)tan(π/6) – ½ = 0
cos(5π/6)tan(5π/6) – ½ = 0
sec(x)cot(x) + 2 =
0 → 7π/6 and 11π/6
sec(7π/6)cot(7π/6) + 2 = 0
sec(11π/6)cot(11π/6) + 2 = 0
sin(x)cot(x) +
1/sqrt2 = 0 → 3π/4 and 5π/4
sin(3π/4)cot(3π/4) + 1/sqrt2 = 0
sin(5π/4)cot(5π/4) + 1/sqrt2 = 0
csc(x)tan(x) – 2 = 0 → π/3 and 5π/3
csc(π/3)tan(π/3) – 2 = 0
csc(5π/3)tan(5π/3) – 2 = 0
