1) The graph consists of three horizontal segments, with discontinuities (jumps) at x = 1, x = 2, and x = 3.
A horizontal segment at y = - 2 for the values x = 0 to 1.
A horizontal segment at y = - 1 for values x = 1 to 2
A horizontal segment at y = 0 for values x - 2 to 3.
2) To know whether the end points of a segment are defined by the left or the right values you have to look for the circle at the extreme of the segment: if it is a solid dot, means that the end is included, if is is an open circle (white inside) then the end is not included in that segment.
3) That function is based on the function named integer part because if relates y with the integer part of x.
The integer value function is [x] and it makes correspond y values witht he integer values of x.:
y = 0 witht the integer value of x for x between 0 and 1, excluding 1.
y = 1 with the integer value of x between 1 and 2 (excluding 2)
y = 2 with the integer value of x between 2 and 3 (excluding 3)
y = 3 with the integer value of x between 3 and 4 (excluding 4)
But our function is two units below, so it is [x] - 2
You would have to add 47.50 + 122 to get 169.5 dollars. to get the amount of money he made in the summer, you would have to multiply 169.5 by 2 which would equal 339 dollars. so he made 339 dollars over the summer :)
Answer:
B
Step-by-step explanation:
we are given a equation of a line
we want to figure out the equation of the perpendicular line passes through the <u>(</u><u>6</u><u>,</u><u>5</u><u>)</u><u> </u>points
in order to do so
recall that,

we got from our given equation that m=2
because equation of a line is y=mx+b
thus

remember that, when we want to figure out perpendicular line or parallel line we should the formula given by

since we got our perpendicular m is -½,
and
, substitute

to get the perpendicular equation you should simplify the above equation to y=mx+b form
distribute -½:

add 5 to both sides:

hence,
our answer choice is B
the standard form of a quadratic formula is
y = ax^2 + bx + c
in this case you will solve using foil method
(× - 4)(x + 3)
<em>(</em><em>x</em><em> </em><em>×</em><em> </em><em>x</em><em>)</em><em> </em><em>+</em><em>(</em><em> </em><em>x</em><em> </em><em>×</em><em> </em><em>3</em><em> </em><em>)</em><em>(</em><em>-</em><em> </em><em>4</em><em> </em><em>×</em><em> </em><em>x</em><em>)</em><em> </em><em>(</em><em> </em><em>-4</em><em>)</em><em>×</em><em> </em><em>3</em><em>)</em><em>)</em>
<em>x</em><em>^</em><em>2</em><em> </em><em>+</em><em> </em><em>3x</em><em> </em><em>-</em><em> </em><em>4x</em><em> </em><em>-12</em>
<em>x</em><em>^</em><em>2</em><em> </em><em>-</em><em> </em><em>x</em><em> </em><em>-</em><em> </em><em>1</em><em>2</em>
<em>therefore</em><em> </em>
<em>y</em><em> </em><em>=</em><em> </em><em>x^</em><em>2-</em><em> </em><em>x</em><em> </em><em><u>-</u></em><em><u> </u></em><em><u>1</u></em><em><u>2</u></em><em><u> </u></em>