1. I understood that "2" in writing "f(x)=log2x" it's a the base of a logarithm.2. I understand that this equation must be written like this "2^(2y)=4", because else the problem lose the meaning.So, for the first, we must to remember what is a logarithm?Logarithm - a number that shows how many times a base number (such as ten) is multiplied by itself to produce a third number (such as 100).In simple terms: Logarithm it's the DEGREE that we must use for a base, to recieve argument.Now let's look at our equation.2^(2y)=4, let's make some changes.By property of degrees:2^(2y)=(2^2)^y.(2^2)^y=44^y=4y=log_{4}4
For find the results in the graph, we need to calculate the value of argument, wich we will search in graph.y=los_{2}x For use this graph we need to change our equation, to bring the equation to the form with such base. For that, we will use the properties of logarithms.log_{4}4=Log_{(2^2)}4=Log_{2}4^(1/2)=log_{2}2.Now for recieve the value of Y, we must to find "2" in axis OX, to build the line at right angle (from axis) into the graph. After that, we turn to axis OY at right angle, and we make the line to axis OY. In result we have the value on the axis OY, it's will be the desired value. And it must be 1.
Answer:
f(2)<u> < </u>g(3)
The correct option is;
1. <
Step-by-step explanation:
The given function parameters are;
f(x) = 3·x - 1, g(x) = 2·x² - 3
For f(x) = 3·x - 1, when x = 2, we have;
f(2) = 3·x - 1 = 3 × 2 - 1 = 5
For g(x) = 2·x² - 3, when x = 3, we have;
g(3) = 2 × 3² - 3 = 18 - 3 = 15
Therefore, we have;
f(2) = 5 < g(3) = 15
Which gives;
f(2) < g(3)
Answer:
6
Step-by-step explanation:
36/6 is 6
<span>There are several possible events that lead to the eighth mouse tested being the second mouse poisoned. There must be only a single mouse poisoned before the eighth is tested, but this first poisoning could occur with the first, second, third, fourth, fifth, sixth, or seventh mouse. Thus there are seven events that describe the scenario we are concerned with. With each event, we want two particular mice to become diseased (1/6 chance) and the remaining six mice to remain undiseased (5/6 chance). Thus, for each of the seven events, the probability of this event occurring among all events is (1/6)^2(5/6)^6. Since there are seven of these events which are mutually exclusive, we sum the probabilities: our desired probability is 7(1/6)^2(5/6)^6 = (7*5^6)/(6^8).</span>
