All you have to do is plug in the given x values. your first equations would read:
f(-3) = 2^(-3)
f(-2) = 2^(-2)
f(-1) = 2^(-1)
these can be solved by moving decimal points or entering them into a calculator. regardless of the method, your answers are:
f(-3) = 0.002
f(-2) = 0.02
f(-1) = 0.2
so just repeat that process to fill in the rest of your table. to graph it, you'll use them as normal (x, y) points:
(-3, 0.002)
(-2, 0.02)
(-1, 0.2)
the graph might be a little difficult, working with such small values, but precision isn't totally important--0.002 will be super close to 0, 0.02 will be slightly further, 0.2 will be slightly further. the smaller values don't matter as much graphically and you'll recognize the graph of a growing exponential as you graph more of the table.
Answer:
The arrow is at a height of 48 ft after approximately <u>0.55</u> s and after <u>5.45</u> s
Step-by-step explanation:
The following information is missing:
<em>The height of an arrow shot upward can be given by the formula s = v0*t - 16*t², where v0 is the initial velocity and t is time.How long does it take for the arrow to reach a height of 48 ft if it has an initial velocity of 96 ft/s? </em>
If the arrow is at a height of 48 ft and its initial velocity is 96 ft/s, then:
48 = 96*t - 16*t²
16*t² - 96*t + 48 = 0
16*(t² - 6*t + 3) = 0
t² - 6*t + 3 = 0
t² - 6*t + 3 + 6 = 0 + 6
t² - 6*t + 9 = 6
(t - 3)² = 6
t - 3 = √6
t - 3 = 2.45; t = 2.45 + 3; t = 5.45
or
t - 3 = -2.45; t = -2.45 + 3; t = 0.55
Firstly, we will find smallest value and largest value
smallest value is in 2009
and smallest value is 3
largest value is in 2010
and largest value is 12
So, percentage change will be greatest between 2009 to 2010
now, we can find percentage change
%..............Answer
Answer:
Step-by-step explanation:
Using the hint, write a and b in the following prime factorization:
where 
for i ≠ j.
Then by the formulae for gcd(a,b) and lcm(a,b) we know that:
Note that the expression 
for all i, since if the minimum is, <em>without loss of generality</em>, 
, then the maximum must be 
, and viceversa. Then, it is straightforward to verify that when we multiply gcd(a, b) and lcm(a, b) its prime factorization matches the prime factorization of ab, and so we can see the equaility holds:
