On what intervals is the function decreasing? Indicate intervals on the x-axis using the Ray tool for intervals that extend to i
nfinity and the Segment tool for bounded intervals. Graph a ray by selecting the endpoint followed by a point on the ray. Graph a segment by selecting an endpoint followed by the other endpoint.
1 answer:
Answer:
( -6,-2 ) and ( 3,7 )
Step-by-step explanation:
We are given the graph of function which is increasing in some parts and decreasing in other.
Now in x < -6, the function is increasing to reach the point ( -6,5 ).
Also, in interval ( -6,-2 ) the function is decreasing to reach the point ( -2,-7 ).
Further, in interval ( -2,3 ) the function is increasing to reach the point ( 3,6 ).
Then, in interval ( 3,7 ) the function is decreasing to reach the point ( 7,-5 ).
Finally, in x > 7, the function is increasing to infinity.
Hence, we see that the intervals where the function is decreasing are ( -6,-2 ) and ( 3,7 ).
Further, the ray AB and the line segment AC plotted in the graph can be seen in the figure below.
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Answer:
1.
2.
a)
35 cm = 0.35 meters
11 dm = 1.1 meters
15 mm = 0.015 meters
b)
Volume = 0.005775 cubic meters
Mass = 15.5925 kilograms
Step-by-step explanation:
1.
We know
Density = Mass/Volume
Given
Mass = 90 kg
Volume = 0.075 cubic meters
The SI unit for density is kilograms per cubic meters. Our dimensions are just that, so we just need to plug in the numbers into the formula and get our answer. Shown below:
Density =
The Density of asphalt block =
2.
a)
We need to express 35cm, 11dm, and 15mm in meters
We know
100 cm = 1 meters, so
35 cm = 35/100 = 0.35 meters
We know 1 dm = 0.1 meters, so
11 dm * 0.1 = 1.1 meters
We know, 1mm = 0.001 meters, so
15 mm * 0.001 = 0.015 meters
b)
Volume of the Slab is length * width * height, in meters, that would be:
Volume = 0.35 * 1.1 * 0.015 = 0.005775 cubic meters
THe mass would be found by the formula:
Density = Mass/Volume
2700 = Mass/0.005775
Mass = 0.005775 * 2700 = 15.5925 kilograms
We know that
case a)
the equation of the vertical parabola write in vertex form is
y=a(x-h)²+k,
where (h, k) is the vertex.
Using our vertex, we have:
y=a(x-2)²-1
We know that the parabola goes through (5, 0),
so
we can use these coordinates to find the value of a:
0=a(5-2)²-1
0=a(3)²-1
0=9a-1
Add 1 to both sides:
0+1=9a-1+1
1=9a
Divide both sides by 9:
1/9 = 9a/9
1/9 = a
y=(1/9)(x-2)²-1
the answer is
a=1/9
case b)
the equation of the horizontal parabola write in vertex form is
x=a(y-k)²+h,
where (h, k) is the vertex.
Using our vertex, we have:
x=a(y+1)²+2,
We know that the parabola goes through (5, 0),
so
we can use these coordinates to find the value of a:
5=a(0+1)²+2
5=a+2
a=5-2
a=3
x=3(y+1)²+2
the answer is
a=3
see the attached figure
case a)
the equation of the vertical parabola write in vertex form is
y=a(x-h)²+k,
where (h, k) is the vertex.
Using our vertex, we have:
y=a(x-2)²-1
We know that the parabola goes through (5, 0),
so
we can use these coordinates to find the value of a:
0=a(5-2)²-1
0=a(3)²-1
0=9a-1
Add 1 to both sides:
0+1=9a-1+1
1=9a
Divide both sides by 9:
1/9 = 9a/9
1/9 = a
y=(1/9)(x-2)²-1
the answer is
a=1/9
case b)
the equation of the horizontal parabola write in vertex form is
x=a(y-k)²+h,
where (h, k) is the vertex.
Using our vertex, we have:
x=a(y+1)²+2,
We know that the parabola goes through (5, 0),
so
we can use these coordinates to find the value of a:
5=a(0+1)²+2
5=a+2
a=5-2
a=3
x=3(y+1)²+2
the answer is
a=3
see the attached figure