To find the surface area, use this square pyramid surface area formula: Square Pyramid Surface Area = 2 x B x S + B 2 B = Width of the Square Base S = Slant length of one of the triangular faces and is calculated from the height and base width by using the equation: S= The square root of [(.5B) 2 + Height 2] .
Answer:
a)
a1 = log(1) = 0 (2⁰ = 1)
a2 = log(2) = 1 (2¹ = 2)
a3 = log(3) = ln(3)/ln(2) = 1.098/0.693 = 1.5849
a4 = log(4) = 2 (2² = 4)
a5 = log(5) = ln(5)/ln(2) = 1.610/0.693 = 2.322
a6 = log(6) = log(3*2) = log(3)+log(2) = 1.5849+1 = 2.5849 (here I use the property log(a*b) = log(a)+log(b)
a7 = log(7) = ln(7)/ln(2) = 1.9459/0.6932 = 2.807
a8 = log(8) = 3 (2³ = 8)
a9 = log(9) = log(3²) = 2*log(3) = 2*1.5849 = 3.1699 (I use the property log(a^k) = k*log(a) )
a10 = log(10) = log(2*5) = log(2)+log(5) = 1+ 2.322= 3.322
b) I can take the results of log n we previously computed above to calculate 2^log(n), however the idea of this exercise is to learn about the definition of log_2:
log(x) is the number L such that 2^L = x. Therefore 2^log(n) = n if we take the log in base 2. This means that
a1 = 1
a2 = 2
a3 = 3
a4 = 4
a5 = 5
a6 = 6
a7 = 7
a8 = 8
a9 = 9
a10 = 10
I hope this works for you!!
Remember that the radicand (the area under the root sign) must be positive or zero for a radical with an even index (like the square root or fourth root, for example). This is because two numbers squared or to the fourth power, etc. cannot be negative, so there are no real solutions when the radicand is negative. We must restrict the domain of the square-root function.
If the domain has already been restricted to
, we can work backwards to add 11 to both sides. We see that
must be under the radicand, so the answer is
A.
Answer:
32 days.
Step-by-step explanation:
56 / 1.75 = 56 / (7/4) = 56 * 4 / 7 = 32.
