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m_a_m_a [10]
3 years ago
6

What is 17.445 - 6.76 rounded to the nearest tenth

Mathematics
1 answer:
Lilit [14]3 years ago
6 0

Answer:

10.7

Step-by-step explanation:

17.445 - 6.76= 10.685

10.685 is 10.7 when rounded to the nearest tenth!

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need it quick 13 points
motikmotik

Answer:

4 And -4 i think it depends

Step-by-step explanation:

3 0
3 years ago
I need help with this question
bearhunter [10]

Answer:

  7.28 miles

Step-by-step explanation:

Suppose the distance at closest approach is represented by x. Then the distance to the point of closest approach at the first sighting is ...

  d1 = x·tan(62°)

At the second sighting, the distance to the point of closest approach is ...

  d2 = x·tan(38°)

The difference of these distances is 8 miles, so we have ...

  d1 -d2 = 8 = x(tan(62°) -tan(38°))

Dividing by the coefficient of x, we find ...

  x = 8/(tan(62°) -tan(38°)) ≈ 7.2764 . . . . miles

The point of closest approach is about 7.28 miles from the landmark.

5 0
3 years ago
Please solve this problem
sladkih [1.3K]

Answer:

Choice C

Step-by-step explanation:

1/2 x < 4    Isolate x by multiplying both sides by 2:-

x < 8  


4 0
3 years ago
Give the number to which the Fourier series converges at a point of discontinuity of f.
quester [9]

Answer:

The Fourier series of f(x) converges to 3 at the points x= π+2kπ, where k is an integer.

Step-by-step explanation:

First, recall that the function f(x) is extended 2π periodic to the whole real line, in order to obtain a valid Fourier expansion. Remember that a Fourier series is formed by a sines and cosines, which are 2π-periodic.

So, the 2π-periodic expansion of f(x) is discontinuous at the points π+2kπ, in particular π and -π. Check the attached figure to a better understanding.

Now, the Dirichlet theorem on the convergence of a Fourier series tells us that the series converges to the function at the points of continuity, and at points of discontinuity the sum of the series is

\frac{f(x_0+)+f(x_0-)}{2}.

Here we understand the notation f(x+) and f(x-) as

f(x_0+) = \lim_{x\rightarrow x_0}f(x), x>x_0

and

f(x_0-) = \lim_{x\rightarrow x_0}f(x), x.

In this particular case

f(\pi-) = \lim_{x\rightarrow \pi}(3-2x) = 3-2\pi, x.

For the limit f(\pi+) = \lim_{x\rightarrow \pi}(3-2x), with x>\pi recall that our function is 2π-periodic, so the values of f near π, with x>π are the same when x is near -π and x>-π. Again, check the attached figure. So,

f(\pi+) = \lim_{x\rightarrow \pi}(3-2x) = 3+2\pi, x.

Thus,

\frac{f(\pi+)+f(\pi-)}{2} = \frac{3+2\pi +3-2\pi}{2} = \frac{6}{2} =3.

Note: In the attached figure we only have drawn three repetitions of the 2π-periodic extension of f, recall that the extension is <em>ad infinitum</em>. Also, the points drawn in the dotted lines are the sum of the series at the points of discontinuity.

6 0
3 years ago
Evaluate the expression for the given values. (5 x - 4 y) 2 given x = 1 and y = -1.
Reika [66]

((5 \times 1) - (4 \times  - 1)  =  \\ 5 - ( - 4) =  \\ 5 + 4 =  \\ 9

3 0
3 years ago
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