Which stage of the business cycle is an economy in when it reaches its low point? Expansion Peak Contraction Trough
1 answer:
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The approximations T8 and M8 for the given integral are:
- T8 = 33.386321; and
- M8 = 33.50794
An Integral is a variable of which a given function is the derivative, i.e. it gives that function when differentiated and may express the area under the curve of the function's graph.
<h3>What is the explanation to above answer?</h3>Given:
F(x) = 37 cox (x²)
Internal = [0,1] n = 8 in Δ x = 1/8
The sub intervals are:
[0, 1/8], [1/8, 2/8], [2/8, 3/8], [ 3/8, 4/8], [ 4/8, 5/8], [ 5/8, 6/8], [6/8, 7/8], [ 7/8, 1]
The mid points are given as:
1/16, 3/16, 5/16, 7/16, 9/16, 11/16, 13/16, 15/16
and X₀ = 0, X₁ = 1/8, X₂ = 2/8
Using the Trapezium Rule which states that: = Δx/2
= 1/1Q[f(0) + 2f (1/8) + 2f(2/8) + ....+ 2f(7/8) + f(1)]
= 0.902333
Now
T8 =
=
= 37 (0.902333)
T8 = 33.386321
It is to be noted that the midpoints rule is given as; = Δx [f(1/16) + (3/16) + .... + f(15/16)]
= 1/8[f(1/16) + f (3/16) + f(5/16) + f(7/16) + f(9/16) + f(11/16) + f(13/16) + f(15/16)]
= 0.905620
From the above,
M8 =
=
= 37 (0.905620)
M8 = 33.50794
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Answer:
3 times
Explanation:
When the dough is folded, it increases by a constant factor. We can model the growth of the thickness using the exponential growth model
Where:
Initial thickness, = 2mm
Growth factor, r =8%=0.08
We want to find the smallest number of times Soon Yi will have to roll and fold the dough so that the resulting dough is at least 2.5mm.
i.e When
Therefore, the smallest number of times Soon Yi will have to roll and fold the dough so that the resulting dough is at least 2.5mm thick is 3.