Both functions decrease at (-1, 2)
<h3>How to determine the decreasing intervals of the function?</h3>
The complete question is added as an attachment
The polynomial function f(x) is represented by the graph.
From the graph, the polynomial function decreases at (-2, 2)
The absolute function is given as;
f(x) = =5|x + 1| + 10
The vertex of the above function is
Vertex = (-1, 10)
Because a is negative (a= -5), the vertex is a maximum.
This means that the function decreases at (-1, ∞)
So, we have
(-2, 2) and (-1, ∞)
Combine both intervals
(-1, 2)
This means that both functions decrease at (-1, 2)
See attachment 2 for the number line that represents the interval where both functions are decreasing
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Answer:
1. The right triangles pass the AA similarity congruence theorem.
2. s/q
Step-by-step explanation:
They are similar since they pass the AA( Angle-Angle) similarity theorem.
Similar triangles have corresponding proportional sides so side z corresponds with side s and side x corresponds with q so we would represent that as
s/q
Answer:
The domain is -7, 0, and 5
The range is -1, 0, and 8
Step-by-step explanation:
The domain of a set of points is the x-value. In this case the x-values are -7, 0, and 5 respectively. The range of a set of points is the y-value so in this case the range is -1, 0, and 8.
Answer:
- 0.944
Step-by-step explanation:
We know that,
cos 3A = 4 cos³A - 3 cos A (formula)
So,
If cos A = 0.4,
cos 3A = 4 cos³A - 3 cos A
cos 3A = 4 × (0.4)³ - 3 × (0.4)
cos 3A = 4 × 0.064 - 1.2
cos 3A = - 0.944
Hope it helps ⚜
The actual width of the room is 20 ft and the actual length of the room is 15 ft
Since on the scale, 2 in : 5 ft and the width of the room on the drawing is 8 in.
Let the actual width of the room is w.
The ratio of the drawing to actual width is 8 in : w
So, 2 in : 5 ft = 8 in : w
2 in/5 ft = 8 in/w
So, w = 8 in × 5 ft/2 in
w = 4 × 5 ft
w = 20 ft
Also, the length of the room on the drawing on the drawing is 6 in.
Let the actual length of the room is L.
The ratio of the drawing to actual length is 6 in : L
So, 2 in : 5 ft = 6 in : L
2 in/5 ft = 6 in/L
So, L = 6 in × 5 ft/2 in
L = 3 × 5 ft
L = 15 ft
So, the actual width of the room is 20 ft and the actual length of the room is 15 ft.
Learn more about scale drawing here:
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