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telo118 [61]
3 years ago
14

Which is the graph of y = ⌊x⌋ – 2? On a coordinate plane, a step graph has horizontal segments that are each 1 unit long. The le

ft end of each segment is a closed circle. The right end of each segment is an open circle. The left-most segment goes from (negative 5, negative 5) to (negative 4, negative 5). Each segment is 1 unit higher and 1 unit farther to the right than the previous segment. The right-most segment goes from (4, 4) to (5, 4). On a coordinate plane, a step graph has horizontal segments that are each 1 unit long. The left end of each segment is an open circle. The right end of each segment is a closed circle. The left-most segment goes from (negative 5, negative 5) to (negative 4, negative 5). Each segment is 1 unit higher and 1 unit farther to the right than the previous segment. The right-most segment goes from (4, 4) to (5, 4). On a coordinate plane, a step graph has horizontal segments that are each 1 unit long. The left end of each segment is a closed circle. The right end of each segment is an open circle. The left-most segment goes from (negative 3, negative 5) to (negative 2, negative 5). Each segment is 1 unit higher and 1 unit farther to the right than the previous segment. The right-most segment goes from (4, 2) to (5, 2). On a coordinate plane, a step graph has horizontal segments that are each 1 unit long. The left end of each segment is an open circle. The right end of each segment is a closed circle. The left-most segment goes from (negative 4, negative 5) to (negative 3, negative 5). Each segment is 1 unit higher and 1 unit farther to the right than the previous segment. The right-most segment goes from (4, 3) to (5, 3).
Mathematics
1 answer:
antoniya [11.8K]3 years ago
8 0

Answer:

On a coordinate plane, a step graph has horizontal segments that are each 1 unit long. The left end of each segment is a closed circle. The right end of each segment is an open circle. The left-most segment goes from (negative 3, negative 5) to (negative 2, negative 5). Each segment is 1 unit higher and 1 unit farther to the right than the previous segment. The right-most segment goes from (4, 2) to (5, 2).

Step-by-step explanation:

The floor function graphs as horizontal segments 1 unit long, each 1 unit up from the segment to its left. It will have a closed circle at the left end of the segment, and an open circle at the right end.

Since 2 is subtracted from the floor of the x-value, the closed circle at the left end of the segment will have coordinates (x, x-2). The only offered choice meeting that condition is the 3rd choice listed here.

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Step-by-step explanation:

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A rectangular garden measures 25ft by 38ft. Surrounding (and bordering) the garden is a path 2ft wide. Find the area of this pat
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The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg
julia-pushkina [17]

Answer:

Approximately 101\; \rm ft (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

  • Let \rm O denote the observer.
  • Let \rm A denote the top of the tree.
  • Let \rm R denote the base of the tree.
  • Let \rm B denote the point where line \rm AR (a vertical line) and the horizontal line going through \rm O meets. \angle \rm B\hat{A}R = 90^\circ.

Angles:

  • Angle of elevation of the base of the tree as it appears to the observer: \angle \rm B\hat{O}R = 51^\circ.
  • Angle of elevation of the top of the tree as it appears to the observer: \angle \rm B\hat{O}A = 57^\circ.

Let the length of segment \rm BR (vertical distance between the base of the tree and the base of the hill) be x\; \rm ft.

The question is asking for the length of segment \rm AB. Notice that the length of this segment is \mathrm{AB} = (x + 20)\; \rm ft.

The length of segment \rm OB could be represented in two ways:

  • In right triangle \rm \triangle OBR as the side adjacent to \angle \rm B\hat{O}R = 51^\circ.
  • In right triangle \rm \triangle OBA as the side adjacent to \angle \rm B\hat{O}A = 57^\circ.

For example, in right triangle \rm \triangle OBR, the length of the side opposite to \angle \rm B\hat{O}R = 51^\circ is segment \rm BR. The length of that segment is x\; \rm ft.

\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}.

Rearrange to find an expression for the length of \rm OB (in \rm ft) in terms of x:

\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}.

Similarly, in right triangle \rm \triangle OBA:

\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}.

Equate the right-hand side of these two equations:

0.810\, x \approx 0.649\, (x + 20).

Solve for x:

x \approx 81\; \rm ft.

Hence, the height of the top of this tree relative to the base of the hill would be (x + 20)\; {\rm ft}\approx 101\; \rm ft.

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Step-by-step explanation:

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