Answer:
y=-2
Step-by-step explanation:
Answer:
Non-mutually exclusive
Step-by-step explanation:
both events can happen simultaneously... 3 is an odd number, events A and B have a common outcome.
It would be mutually exclusive if event A was rolling a 4 and event B was rolling an odd number.
Answer:
x = 5
Step-by-step explanation:
6(3 - 7x) = -192
Divide each side by 6
6/6(3 - 7x) = -192/6
3 - 7x = -32
Subtract 3 from each side
3-7x-3 = -32-3
-7x = -35
Divide each side by -7
-7x/-7 = -35/-7
x = 5
A key feature is there is a constant y-value level between x values from 0 to 15 and there is a parabolic curve from x-values of 15 to 65. The vertex is at (3, 45)
<h3>How to get the relationship between graphs?</h3>
A) This is a parabolic graph and from the graph, we see that, the y-values remain the same from x-values of 0 to 15. Thereafter the x-values increases with a corresponding decrease in y-values until the vertex point before increase in x-values with corresponding increase in y-values.
A relationship could be: Battery percentage remains at the mark of 40 for the first 15 minutes of use. Thereafter, it begins to decrease parabolically until 45 minutes when it it is almost at 0 level and is charged before it starts to increase in a parabolic manner again for another 20 minutes when it increases linearly.
B) A key feature is there is a constant y-value level between x values from 0 to 15 and there is a parabolic curve from x-values of 15 to 65. The vertex is at (3, 45)
Read more about Graph relationships at; brainly.com/question/13060180
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Answer:
See proof below
Step-by-step explanation:
An equivalence relation R satisfies
- Reflexivity: for all x on the underlying set in which R is defined, (x,x)∈R, or xRx.
- Symmetry: For all x,y, if xRy then yRx.
- Transitivity: For all x,y,z, If xRy and yRz then xRz.
Let's check these properties: Let x,y,z be bit strings of length three or more
The first 3 bits of x are, of course, the same 3 bits of x, hence xRx.
If xRy, then then the 1st, 2nd and 3rd bits of x are the 1st, 2nd and 3rd bits of y respectively. Then y agrees with x on its first third bits (by symmetry of equality), hence yRx.
If xRy and yRz, x agrees with y on its first 3 bits and y agrees with z in its first 3 bits. Therefore x agrees with z in its first 3 bits (by transitivity of equality), hence xRz.
