1. 11; 4
2. -64; 81
3. X+5; x+22,500
4. -13; -3
5. 8
6. 56F; 6F
7. -29/10 or -2 and 9/10
8. >; <; <
9. -2/9
Answer:
2/11
Step-by-step explanation:
Let's say our fraction is x = 0.1818181818...
The trick is to multiply x by 10²=100 in this case, since there are two repeating digits, and then subtract the original x.
So, in fact you are subtracting 0.181818 from 18.181818 which effectively cancels the entire bit after the decimal point.
You get:
100x - x = 18
Which you can solve:
99x = 18
x = 18/99 = 2/11
9514 1404 393
Answer:
779.4 square units
Step-by-step explanation:
You seem to have several problems of this type, so we'll derive a formula for the area of an n-gon of radius r.
One central triangle will have a central angle of α = 360°/n. For example, a hexagon has a central angle of α = 360°/6 = 60°. The area of that central triangle is given by the formula ...
A = (1/2)r²sin(α)
Since there are n such triangles, the area of the n-gon is ...
A = (n/2)r²sin(360°/n)
__
For a hexagon (n=6) with radius 10√3, the area is ...
A = (6/2)(10√3)²sin(360°/6) = 450√3 ≈ 779.4 . . . . square units
Answer:
Graph A → y=√x.
Graph B → y=(√x) - 1.
Graph C → y=√(x-1).
Graph D → y= -√x.
Graph E → y= -√(x-1)
Step-by-step explanation:
The graph 'A' intercepts the y-axis at (0, 0). Therefore it belongs to the function y=√x.
The graph 'D' is exactly the same graph 'A' but reflected across the x-axis. Therefore, it belongs to the function y=-√x.
The function 'C' is exactly the same function y=√x but translated one unit to the right, therefore, the solution function is y=√(x-1)
The graph 'E' is exactly the same graph 'C' but reflected across the x-axis, therefore the function is: y= -√(x-1)
In the options you have two times the function y=√x. I assume that's a mistake. The graph 'B' corresponds to y = (√x) - 1
In the 
direction we consider the 
subintervals [0, 1] and [1, 2] (each with length 1), while in the 
direction we consider the 
subintervals [0, 2] and [2, 4] (with length 2). Then the lower right corners of the cells in the partition of 
are (1, 0), (2, 0), (1, 2), (2, 2).
Let 
. The volume of the solid is approximately
###
More generally, the lower-right-corner Riemann sum over 
and 
subintervals would be
Then taking the limits as 
and 
leaves us with an exact volume of 
.
