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2 years ago

Approximately 10% of all people are left-handed. Consider a grouping of fifteen people.

Mathematics

1 answer:

vaieri [72.5K]2 years ago

8 0

1.5 but you can’t get half a person

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Find the 20th term of the following sequence. -6, -4, -2, 0, ... 32 44 46

mars1129 [50]

You are adding 2 each time.

4 terms are given to you, with 5th term being 2

20 - 5 = 15

Multiply 15 with 2

15 x 2 = 30

Add the number gotten with the last number given

30 + 2 = 32

32, or (A) is your answer

hope this helps

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2 years ago

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oee [108]

Answer:

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Step-by-step explanation:

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2 years ago

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The graph of which function is decreasing over the interval (–4, ∞)?

padilas [110]

I think second must b the answee

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2 years ago

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A point is represented by a single dot and

evablogger [386]

A letter, which is how you name it. If that is what your asking, of course.

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2 years ago

Evaluate the integral. (sec2(t) i t(t2 1)8 j t7 ln(t) k) dt

polet [3.4K]

If you're just integrating a vector-valued function, you just integrate each component:

$\displaystyle\int(\sec^2t\,\hat\imath+t(t^2-1)^8\,\hat\jmath+t^7\ln t\,\hat k)\,\mathrm dt$

$=\displaystyle\left(\int\sec^2t\,\mathrm dt\right)\hat\imath+\left(\int t(t^2-1)^8\,\mathrm dt\right)\hat\jmath+\left(\int t^7\ln t\,\mathrm dt\right)\hat k$

The first integral is trivial since $(\tan t)'=\sec^2t$ .

The second can be done by substituting $u=t^2-1$ :

$u=t^2-1\implies\mathrm du=2t\,\mathrm dt\implies\displaystyle\frac12\int u^8\,\mathrm du=\frac1{18}(t^2-1)^9+C$

The third can be found by integrating by parts:

$u=\ln t\implies\mathrm du=\dfrac{\mathrm dt}t$

$\mathrm dv=t^7\,\mathrm dt\implies v=\dfrac18t^8$

$\displaystyle\int t^7\ln t\,\mathrm dt=\frac18t^8\ln t-\frac18\int t^7\,\mathrm dt=\frac18t^8\ln t-\frac1{64}t^8+C$

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2 years ago