To determine the 'intervals of increase' and 'intervals of decrease' we can refer to the graph with respect to the x - axis.
• Knowing that t = x - axis, the 'intervals of increase' as an inequality would be 1 < x < 3, and 4 < x < ∞. Therefore we have our intervals of increase as (1,3) and (4, ∞).
• Respectively our 'intervals of decrease' as inequalities would be - ∞ < x < 1, and 3 < x < 4. Our intervals of decrease would then be (- ∞, 1) and (3,4).
• We are left with our local extrema and absolute extrema. Now remember the absolute extrema is the absolute lowest point in the whole graph, while the local extrema is the lowest point in a restricted interval. In this case our local extrema is our maximum, (3,1). But this maximum is not greater than the starting point (0, 4) so it appears, and hence their is no absolute extrema.
Answer:
f(n)= f(1)+d(n-1)
Step-by-step explanation:
the aerithmetric explict formula is the one above, but im not sure what the question is asking, your difference is 4 btw
Answer:
a) 127m (3 s.f.)
b) 271m (3 s.f.)
Step-by-step explanation:
Please see the attached pictures for full solution.
Method 1 is written in black and method 2 is written in pink.
All we have to do here is to express numbers 40 and 5 in each of equations. Note that first number in brackets is always x coordinate and second number is y coordinate
First equation:
5 = 40*8 false
Second
5 = 5*40 false again
Third:
5 = 40/8 = 5 True
checking for just in case last one:
5 = 40/5 = 8 False
Correct one is C)