We are given with
a1 = 2
r = 4
These are components of a geometric series. The first term is 2 and the common ratio is 4. To get the first six terms, we use the formula:
an = a1 r^(n-1)
a1 = 2 (4)^(1-1) = 2
a2 = 2 (4)^(2-1) = 8
a3 = 2 (4)^(3-1) = 32
a4 = 2 (4)^(4-1) = 128
a5 = 2 (4)^(5-1) = 512
a6 = 2 (4)^(6-1) = 2048
Answer:
10.4900
Step-by-step explanation:
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The area between the two functions is 0
<h3>How to determine the area?</h3>
The functions are given as:
f₁(x)= 1
f₂(x) = |x - 2|
x ∈ [0, 4]
The area between the functions is
A = ∫[f₂(x) - f₁(x) ] dx
The above integral becomes
A = ∫|x - 2| - 1 dx (0 to 4)
When the above is integrated, we have:
A = [(|x - 2|(x - 2))/2 - x] (0 to 4)
Expand the above integral
A = [(|4 - 2|(4 - 2))/2 - 4] - [(|0 - 2|(0 - 2))/2 - 0]
This gives
A = [2 - 4] - [-2- 0]
Evaluate the expression
A = 0
Hence, the area between the two functions is 0
Read more about areas at:
brainly.com/question/14115342
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The answer is the first choice, "5 and 6."
The square root of 35 is equal to 5.92, which is in between the numbers 5 and 6.
