The given quadratic describes a parabola that opens upward. Its one absolute extreme is a minimum that is found at x = -3/2. The value of the function there is
(-3/2 +3)(-3/2) -1 = -13/4
The one relative extreme is a minimum at
(-1.5, -3.25).
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For the parabola described by ax² +bx +c, the vertex (extreme) is found where
x = -b/(2a)
Here, that is x=-3/(2·1) = -3/2.
Answer:
x = e^2/2 + 3
Step-by-step explanation:
Solve for x:
log(2 x - 6) = 2
Hint: | Eliminate the logarithm from the left hand side.
Cancel logarithms by taking exp of both sides:
2 x - 6 = e^2
Hint: | Isolate terms with x to the left hand side.
Add 6 to both sides:
2 x = e^2 + 6
Hint: | Solve for x.
Divide both sides by 2:
Answer: x = e^2/2 + 3
Answer:
219
Step-by-step explanation:
83 + 119 = 202
202 + 17 = 219
16 is correct I hope this helps :)))))))))))))
Answer:
joint variation means L (safe load) is directly proportional to both w and d2
inverse proportionality means L = k/l
So the equation is:
L = k(wd2/l)
You are given:
w = 6 in
d = 5 in
l = 12 ft
L = 7666 lbs
7666 = k(6*52/12)
Solve for the constant, k.
7666 = k (6*25/12)
7666 = k (150/12)
7666 = k (12.5)
613.28 lbs/(ft.in3) = k
Use this k-value to solve for L in the last part of the question.
What safe load, L would a beam 3 in. wide, 7 in. deep and 15 ft long of the same material support? (Round off your answer to the nearest pound.)
w = 3 in
d =7 in
l = 15 ft
k = 613.28 lbs/(ft.in3)
Final Safe load = 6010.144 lbs *** Edited for clarity and to fix a multiplication erro