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Nataliya [291]
2 years ago
8

5) Mary drives 33 miles in 3/5 of an hour. After 6 hours of traveling, 1 point

Mathematics
1 answer:
VMariaS [17]2 years ago
7 0

Answer:

330 miles

Step-by-step explanation:

speed of marry  = distance traveled /time taken

=33/3/5 = 55 miles/hr

therefore distance travelled by marry in 6 hours = 55*6 = 330 miles.

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CALCULUS: For an object whose velocity in ft/sec is given by v(t) = sin(t), what is its distance, in feet, travelled on the inte
rodikova [14]

The linked answer is wrong because that integral gives you the net displacement of the object, not the total distance.

To get the distance, you have to integrate the speed (as opposed to velocity), which involves integrating the absolute value of the velocity function.

\mathrm{distance} = \displaystyle\int_1^5 |\sin(t)| \,\mathrm dt

By definition of absolute value,

|\sin(t)|=\begin{cases}\sin(t)&\text{for }\sin(t)\ge0\\-\sin(t)&\text{for }\sin(t)

Over this particular integration interval,

• sin(<em>t</em> ) ≥ 0 for 1 ≤ <em>t</em> < <em>π</em>, and

• sin(<em>t</em> ) < 0 for <em>π</em> < <em>t</em> ≤ 5

so you end up splitting the integral at <em>t</em> = <em>π</em> as

\mathrm{distance} = \displaystyle\int_1^\pi \sin(t)\,\mathrm dt + \int_\pi^5 (-\sin(t))\,\mathrm dt

Now compute the distance:

\mathrm{distance} = -\cos(t)\bigg|_1^\pi + \cos(t)\bigg|_\pi^5

\mathrm{distance} = -(\cos(\pi) - \cos(1)) + (\cos(5) - \cos(\pi))

\mathrm{distance} = -2\cos(\pi) + \cos(1) + \cos(5) \approx 2.82

making B the correct answer.

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Show the reflection of the given points.Locate the points of the reflection in their proper position around the x-axis
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Can a right triangle be formed using these squares
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2 years ago
Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Viktor [21]

When n=0, we have

5^{3^0} + 1 = 5^1 + 1 = 6

3^{0 + 1} = 3^1 = 3

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for n=k, that

3^{k + 1} \mid 5^{3^k} + 1

Now for n=k+1, we have

5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)

so we know the left side is at least divisible by 3^{k+1} by our assumption.

It remains to show that

3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1

which is easily done with Fermat's little theorem. It says

a^p \equiv a \pmod p

where p is prime and a is any integer. Then for any positive integer x,

5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3

Furthermore,

5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3

which goes all the way down to

5^{3^k} \equiv 5 \pmod 3

So, we find that

\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3

QED

5 0
1 year ago
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