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sdas [7]
4 years ago
14

Heartbeat takes 0.8 seconds.how many seconds is this written as a fraction?

Mathematics
2 answers:
never [62]4 years ago
8 0
0.8 written as a fraction is 4/5
liubo4ka [24]4 years ago
5 0
0.8 = 8/10

simplified = 4/5 seconds

hope this helps
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Find the value of 3x+2 when x=5 ​
babunello [35]

Answer:

17

Step-by-step explanation:

5 0
3 years ago
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Determine whether each of these sets is finite, countably infinite, or uncountable. For those that are countably infinite, exhib
anzhelika [568]

Answer:

a) Countably infinite

b) Countably infinite

c) Finite

d) Uncountable

e) Countably infinite

Step-by-step explanation:

a) Let S the set of integers grater than 10.

Consider the following correspondence:

f: S\rightarrow \mathbb{Z}^+ defined by f(10+k)=k-1 for k\in\mathbb{Z}^+/\{0\}.

Let's see that the function is one-to-one.

Suppose that f(10+k)=f(10+j) for k≠j. Then k-1=j-1. Thus k-j=1-1=0. Then k=j. This implies that 10+k=10+j. Then the correspondence is injective.

b) Let S the set of odd negative integers

Consider the following correspondence:

f: S\rightarrow \mathbb{Z}^+ defined by f(-(2k+1))=k.

Let's see that the function is one-to-one.

Suppose that f(-(2k+1))=f(-(2j+1)) for k≠j. By definition, k=j. This implies that the correspondence is injective.

c) The integers with absolute value less than 1,000,000 are in the intervals A=(-1.000.000, 0) B=[0, 1.000.000). Then there is 998.000 integers in A that satisfies the condition and 999.000 integers in B that satifies the condition.

d) The set of real number between 0 and 2 is the interval (0,2) and you can prove that the interval (0,2) is equipotent to the reals. Then the set is uncountable.

e) Let S the set A×Z+ where A={2,3}

Consider the following correspondence:

f: S\rightarrow \mathbb{Z}^+ defined by f(2,k)=2k, \;f(3,j)= 2j+1

Let's see that the function is one-to-one.

Consider three cases:

1. f(2,k)=f(2,j), then 2k=2j, thus k=j.

2. f(3,k)=f(3,j), then 2k+1=2j+1, then 2k=2j, thus k=j.

3.  f(2,k)=f(3,j), then 2k=2j+1. But this is impossible because 2k is an even number and 2j+1 is an odd number.

Then we conclude that the correspondence is one-to-one.

6 0
3 years ago
Suppose 17.9% of U.S. households are in the Northeast. Suppose 5.4% of U.S.
Tomtit [17]
In USA Dallas Kansas
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3 years ago
Tell which is the better buy. Show your work.
kramer

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7 0
3 years ago
A steady wind blows a kite due west. The kite's height above ground from horizontal position x = 0 to x = 80 ft is given by y =
GaryK [48]

Answer:

The distance travelled by the kite is 122.8 ft ( approx )

Step-by-step explanation:

Here, the given function,

y=150-\frac{1}{40}(x-50)^2

Differentiating with respect to x,

y'=-\frac{1}{20}(x-50)

∵ arc length of a curve is,

L=\int_{a}^{b} \sqrt{1+y'^2}dx

Where, y shows the height of the curve for a ≤ x ≤ b,

Thus, the arc length of the given curve is,

L=\int_{0}^{80} \sqrt{1+(-\frac{1}{20}(x-50)^2}dx

Put -\frac{1}{20}(x-50)=tan\theta

\implies -dx=-20 sec^2\theta d\theta

\implies L=-20\int_{0}^{80} \sqrt{1+tan^2\theta}sec^2\theta d\theta

=-20\int_{0}^{80} (sec \theta ) sec^2\theta d\theta

=-20\int_{0}^{80} (sec \theta ) sec^2\theta d\theta

By integration by parts,

=|-\frac{20}{2}(sec \theta tan\theta +ln|sec\theta +tan\theta |) |^{x=80}_{x=0}

If x = 80, tan \theta = -\frac{1}{20}(30-50)=\frac{3}{2}

sec \theta = \frac{\sqrt{13}}{2}

\implies \theta = \frac{1}{20}(0-50)=\frac{5}{2}

sec \theta = \frac{\sqrt{29}}{2}

Thus, the length of the curve is,

=-10(\frac{\sqrt{13}}{2}(-\frac{3}{2}) +ln|\frac{13}{2}-\frac{3}{2}|) + 10(\frac{5\sqrt{29}}{4} + ln |\frac{29}{2} + \frac{5}{2} |)

\approx 122.8\text{ feet}

8 0
3 years ago
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