Answer:
y=2 is a line parallel to the x axis that passes through the point (0,2)
A line perpendicular to it would any line that is parallel to the y axis, which crosses the x axis at any point.
Step-by-step explanation:
Perpendicular lines are always found by reciprocating the negative value of the slope in question.
The slope in this case, in y = 2, is zero. You have a horizontal line hovering at the value “2” in the y-dimension, parallel to the x dimension. You can tell the slope is zero since there is no coefficient value paired with “x” so you can assume that since value multiplied by 0 is zero, the same instance has been performed in favor of a dearth of “x.”
Alright. So slope zero. The negative of zero is zero, since zero is neutral. It is neither positive nor negative (though some people tend to see it as positive for reasons irrelevant to your question. Will be answered at your request).
If you take the reciprocal of zero, it becomes undefined because what was once a numerator (0 divided by any number except 0, because then it would already be undefined) becomes a denominator (any numerator divided by 0 becomes undefined by default, as explained already). So now you can assume that the new line is undefined. This means that it is vertical! The only line that does not pass as a function. And it makes sense. A slope of 0 is 90° from a slope of no defined value, which will take on an unknown x-value since its location is not specified.
Answer:
x ≈ 14.87
Step-by-step explanation:
Using Pythagoras' identity in the right triangle.
The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is
x² = 10² + 11² = 100 + 121 = 221 ( take the square root of both sides )
x = 
≈ 14.87 ( to the nearest hundredth )
Answer:
43:100
Step-by-step explanation:
this is the simplest form
Answer: velocity-time graph
Step-by-step explanation: The principle is that the slope of the line on a velocity-time graph reveals useful information about the acceleration of the object. If the acceleration is zero, then the slope is zero (i.e., a horizontal line). If the acceleration is positive, then the slope is positive (i.e., an upward sloping line).
