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!!
The answer is Function A. This is because Function A has a higher y-intercept than Function B. #brainliest
P = k/w^2
<span>first of all you need to find k </span>
<span>But I kind of don't like the units of each value so let say </span>
<span>400KPa = 4 * 10^5 Pa = 4 * 10^5 N/m^2 </span>
<span>2 cm = 2 * 10^-2 m </span>
<span>input values in the equation </span>
<span>4*10^5 = k/(4*10^-4) >>> k = 160 N </span>
<span>P = 160/w^2 </span>
<span>it said if she wears a heel with a width of .5 cm </span>
<span>P = 160(0.5 * 10^-2)^2 >>> P = 6400 KPa</span>
Answer: (5, 12)
Step-by-step explanation:
Just graph the linear equations and find where they intersect.
Algebraically, you can set them equal to each other
-3x-3=2x+2
-x-3=2
-x=5
x=5
Plug x=5 to any equation
y=2(5)+2
y=12
