The negative integers greater than or equal to -12
2 answers:
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Answer:
35
Step-by-step explanation:
7 orchids can be lined as 7!. This means that for the first orchid of the line, you can select 7 options. When you place the first orchid, for the second option you can select among 6 since 1 orchid has already been placed. Similarly, for the 3rd orchid of the line, you have left 5 options. The sequence goes in this fashion and for 7 orchids, you have 7*6*5*4*3*2*1 possibilities. However, there is a restriction here. 3 of the orchids are white and 4 are levender. This means that it does not make a difference if we line 3 white orchids in an arbitrary order since it will seem the same from the outside. As a result, the options for lining the 7 orchids diminish. The reduction should eliminate the number of different lining within the same colors. Similar to 7! explanation above, 3 white orchids can be lined as 3! and 4 levender orchids can be lined as 4!. To eliminate these options, we divide all options by the restrictions. The result is:
= 35. [(7*6*5*4*3*2*1/(4*3*2*1*3*2*1)]
Answer:
The absolute value can be graphed using the points around the vertex (-2, 0), (-4, 2), (-3, 1), (-1, 1), (0, 2).
Step-by-step explanation:
The find the x coordinate of the vertex, set the inside of the absolute value x + 2 equal to 0. In this case, x + 2 = 0.
x + 2 = 0
Subtract 2 from both sides of the equation.
x = -2
Replace the variable x with -2 in the expression.
y = |(-2) + 2|
The absolute value vertex is (-2, 0).
The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.
For each x value, there is on y value. Select few z values from the domain. It would be more useful to select the values so that they are around the x value of the absolute value vertex.
Substitute the x value -4 into f(x) = |x+2|. In this case, the point is (-4,2).
y=2
Substitute the x value -3 into f(x) = |x+2|. In this case, the point is (-3,1).
y=1
Substitute the x value 0 into f(x) = |x+2|. In this case, the point is (0,2).
y=2
You can find the graph in the attachment.
