The domain of the function f(x) = √2-x is the set of numbers x satisfying:
1 answer:
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The solution is the point of intersection between the two equations.
Assuming you have a graphing calculator or a program to lets you graph equations (I use desmos) you simply put in the equetions and note down the coordinates of the point of intersection.
In the graph the first equation is in blue and the second in red.
The point of intersection = the solution = (-6 , -1)
If you dont have access to a graphing calculator you could draw the graphs by hand;
1) Draw a table of values for each equation; you do this by setting three or four values for x and calculating its image in y (you can use any values of x)
y = 0.5 x + 2 (Im writing 0.5 instead of 1/2 because I find its easier in this format)
x | y
-1 | 1.5 * y = 0.5 (-1) + 2 = 1.5
0 | 2 * y = 0.5 (0) + 2 = 2
1 | 2.5 * y = 0.5 (1) + 2 = 2.5
2 | 3 * y = 0.5 (2) + 2 = 3
y = x + 5
x | y
-1 | 4 * y = (-1) + 5 = 4
0 | 5 * y = (0) + 5 = 5
1 | 6 * y = (1) + 5 = 6
2 | 7 * y = (2) + 5 = 7
2) Plot these point on the graph
I suggest to use diffrent colored points or diffrent kinds of point markers (an x or a dot) to avoid confusion about which point belongs to which graph
3) Using a ruler draw a line connection all the dots of one graph and do the same for the other
4) The point of intersection is the solution
Assuming you have a graphing calculator or a program to lets you graph equations (I use desmos) you simply put in the equetions and note down the coordinates of the point of intersection.
In the graph the first equation is in blue and the second in red.
The point of intersection = the solution = (-6 , -1)
If you dont have access to a graphing calculator you could draw the graphs by hand;
1) Draw a table of values for each equation; you do this by setting three or four values for x and calculating its image in y (you can use any values of x)
y = 0.5 x + 2 (Im writing 0.5 instead of 1/2 because I find its easier in this format)
x | y
-1 | 1.5 * y = 0.5 (-1) + 2 = 1.5
0 | 2 * y = 0.5 (0) + 2 = 2
1 | 2.5 * y = 0.5 (1) + 2 = 2.5
2 | 3 * y = 0.5 (2) + 2 = 3
y = x + 5
x | y
-1 | 4 * y = (-1) + 5 = 4
0 | 5 * y = (0) + 5 = 5
1 | 6 * y = (1) + 5 = 6
2 | 7 * y = (2) + 5 = 7
2) Plot these point on the graph
I suggest to use diffrent colored points or diffrent kinds of point markers (an x or a dot) to avoid confusion about which point belongs to which graph
3) Using a ruler draw a line connection all the dots of one graph and do the same for the other
4) The point of intersection is the solution
Answer:
Step-by-step explanation:
Total number of toll-free area codes = 6
A complete number will be of the form:
800-abc-defg
Where abcdefg can be any 7 numbers from 0 to 9. This holds true for all the 6 area codes.
Finding the possible toll free numbers for one area code and multiplying that by 6 will give use the total number of toll free numbers for all 6 area codes.
Considering: 800-abc-defg
The first number "a" can take any digit from 0 to 9. So there are 10 possibilities for this place. Similarly, the second number can take any digit from 0 to 9, so there are 10 possibilities for this place as well and same goes for all the 7 numbers.
Since, there are 10 possibilities for each of the 7 places, according to the fundamental principle of counting, the total possible toll free numbers for one area code would be:
Possible toll free numbers for 1 area code = 10 x 10 x 10 x 10 x 10 x 10 x 10 =
Since, there are 6 toll-free are codes in total, the total number of toll-free numbers for all 6 area codes =
Probabilities are used to determine the chances of selecting a kind of donut from the box.
The probability that Warren eats a chocolate donut, and then a custard filled donut is 0.068
The given parameters are:
The total number of donuts in the box is:
The probability of eating a chocolate donut, and then a custard filled donut is calculated using:
So, we have:
Simplify
Multiply
Divide
Hence, the probability that Warren eats a chocolate donut, and then a custard filled donut is approximately 0.068
Read more about probabilities at: