The correct answer for this question is this one: "B. increase your scale values"
<span>When creating a scatterplot, if the points are too close together to see the relationship, You adjust your graph by </span><em>increasing your scale values</em>
Hope this helps answer your question and have a nice day ahead.
Answer:
Blue splat
Step-by-step explanation:
Blue splat
Answer:
Please find attached the graph of the following function;

Step-by-step explanation:
We note that the function is linear from x = 2 to just before x = 0
The linear relationship of the function f(x) with x changes just before x = 0
At x = 0, the value of f(x) is indicated as 1
From just after x = 0, the function is a straight horizontal line y = 3
The function also changes value immediately after x = 0 to the line y = 3
The areas where the function is defined are shown in continuous lines
Answer:
a)
, b)
, c)
, d) 
Step-by-step explanation:
a) Let assume an initial mass m decaying at a constant rate k throughout time, the differential equation is:

b) The general solution is found after separating variables and integrating each sides:

Where
is the time constant and 
c) The time constant is:


The particular solution of the differential equation is:

d) The amount of radium after 300 years is:
