5. Order from least to greatest. 4996, 10 48% 3. 41 0.408, 7100
2 answers:
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Answer:
Summation notation:
or after using your function part:
After evaluating you get 11.125 square units.
Step-by-step explanation:
The width of each rectangle is the same so we want to take the distance from x=0 to x=3 and divide by 6 since we want 6 equal base lengths for our rectangles.
The distance between x=0 and x=3 is (3-0)=3.
We want to divide that length of 3 units by 6 which gives a length of a half per each base length.
We are doing right endpoint value so I'm going to stat at x=3. The first rectangle will be drawn to the height of f(3).
The next right endpoint is x=3-1/2=5/2=2.5, and the second rectangle will have a height of f(2.5).
The next will be at x=2.5-.5=2, and the third rectangle will have a height of f(2).
The fourth rectangle will have a height of f(2-.5)=f(1.5).
The fifth one will have a height of f(1.5-.5)=f(1).
The last one because it is the sixth one will have a height of f(1-.5)=f(.5).
So to find the area of a rectangle you do base*time.
So we just need to evaluate:
or by factoring out the 1/2 part:
To find f(3) replace x in -x^2+2x+4 with 3:
-3^2+2(3)+4
-9+6+4
1
To find f(2.5) replace x in -x^2+2x+4 with 2.5:
-2.5^2+2(2.5)+4
-6.25+5+4
2.75
To find f(2) replace x in -x^2+2x+4 with 2:
-2^2+2(2)+4
-4+4+4
4
To find (1.5) replace x in -x^2+2x+4 with 1.5:
-1.5^2+2(1.5)+4
-2.25+3+4
4.75
To find f(1) replace x in -x^2+2x+4 with 1:
-1^2+2(1)+4
-1+2+4
5
To find f(.5) replace x in -x^2+2x+4 with .5:
-.5^2+2(.5)+4
-.25+1+4
4.75
Now let's add those heights. After we obtain this sum we multiply by 1/2 and we have our approximate area:
Okay now if you wanted the summation notation for:
is it
or after simplifying a bit:
If you are wondering how I obtain the .5+.5(k-1):
I realize that 3,2.5,2,1.5,1,.5 is an arithmetic sequence with first term .5 if you the sequence from right to left (instead of left to right) and it is going up by .5 (reading from right to left.)
We can see that revolving the region formed by intersecting 3 lines, we will get 2 cones that are connected their bases.
Volume of the cone V=1/3 *πr²*h
1) small cone has r=5, and h=5
Volume small cone V1= 1/3 *π*5²*5 = 5³/3 *π
2) large cone has r=5, and h=21-6=15, h=15
Volume large cone V2= 1/3 *π*5²*15 = 5³*π
3) whole volume
5³/3 *π + 5³*π=5³π(1/3+1)=((5³*4)/3)π=(500/3)π≈166.7π≈523.6
Area
we see 2 right triangles,
Area of the triangle=1/2*b*h, where b -base, h -height
1) small one, b=5, h=5
A1=(1/2)*5*5=25/2
2)large one, b=5, h=15
A2=(1/2)*5*15=75/2
3) whole area=A1+A2=25/2+75/2=100/2=50
Volume of the cone V=1/3 *πr²*h
1) small cone has r=5, and h=5
Volume small cone V1= 1/3 *π*5²*5 = 5³/3 *π
2) large cone has r=5, and h=21-6=15, h=15
Volume large cone V2= 1/3 *π*5²*15 = 5³*π
3) whole volume
5³/3 *π + 5³*π=5³π(1/3+1)=((5³*4)/3)π=(500/3)π≈166.7π≈523.6
Area
we see 2 right triangles,
Area of the triangle=1/2*b*h, where b -base, h -height
1) small one, b=5, h=5
A1=(1/2)*5*5=25/2
2)large one, b=5, h=15
A2=(1/2)*5*15=75/2
3) whole area=A1+A2=25/2+75/2=100/2=50