Integers are whole numbers, both positive and negative. Therefore, all of the positive and negative whole numbers between these two numbers in a set will be your answer:
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It is 104.5
Please mark me the Brainliest answer! Hope this helps. Hugs.
Answer:
- The graph that represents a reflection of f(x) across the x-axis is the blue line on the picture attached.
Explanation:
The function f(x) is:
Which is an exponential function with these features:
- y-intercept: f(0) = 6(0.5)⁰ = 6(1) = 6
- multiplicative rate of change: 0.5 (the base of the exponential term), which means that it is a decaying function (decreasing)
- Horizontal asympote: y = 0 (this is the limit of f(x) when x approaches +∞.
The reflection of f(x) across the x-axis is a function g(x) such that g(x) = - f(x).
Thus, the reflection of f(x) across the x-axis is:
The features of that function are:
- Limit when x approaches - ∞: -∞ (thus the function starts in the third quadrant).
- y-intercerpt: g(0) = -6 (0.5)⁰ = -6(1)= - 6.
- Horizontal asympote: y = 0 (this is the limit of f(x) when x approaches +∞.
- Note that the function never touches the x-axis, thus the function increases from -∞, crosses the y-axis at (0, -6) and continous growing approaching the x-axis but never touchs it. So, this is an increasing frunction, that starts at the third quadrant and ends in the fourth quadrant.
With those descriptions, you can sketch the graph, which you can see in the figure attached. There you have the function f(x) (the red increasing line) and its reflection across the x-axis (the blue increasing line).
Answer:
it's either the 1st one or the 2nd one
Step-by-step explanation:
<span>Answer: To set up the integral, we divide the upper half of the aquarium into horizontal slices,
and for each slice, let x denote its distance from the top of the tank and ∆x denote
2
its thickness. (We choose horizontal slices because we want each drop of water in a
given slice to be the same distance from the top of the tank.) Using the formulae at
the beginning of this handout, we see that the work taken to pump such a slice out of
the tank is
work for a slice = W
= F · d
= (m · a) · d
= (ρ · V ) · a · d .
Since the length, width and thickness of the slice are given by 2 m, 1 m and ∆x m,
respectively, its volume is 2 · 1 · ∆x m3 = 2∆x m3
. Thus, the equation above becomes
work for a slice ≈
force
z }| {
mass
z }| {
(1000 kg/m
3
)
| {z }
density
(2∆x m
3
)
| {z }
volume
(9.8 m/s
2
)
| {z }
gravity
(x m)
| {z }
distance
= (1000)(9.8)(2)x · ∆x (kg · m/s
2
) · m
= (1000)(9.8)(2)x · ∆x N · m
= (1000)(9.8)(2)x · ∆x J .
Summing over our slices, this is
total work for top half of aquarium ≈
X(1000)(9.8)(2)x · ∆x J ,
where the sum is over the slices in the top half of the aquarium; that is, from distance
x = 0 to x = 1/2. As we refine our slices, this becomes the integral
total work = Z 1/2
0
(1000)(9.8)(2)x dx J
= (1000)(9.8)(2) Z 1/2
0
x dx J
= (1000)(9.8)(2)(1/8) J
= 2450 J .</span>