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3 years ago

You pick a card at random what is the probability of picking 8 and then picking 8

Mathematics

2 answers:

dybincka [34]3 years ago

6 0

Answer:

i think 1/9

Step-by-step explanation:

Kruka [31]3 years ago

4 0

Answer:

2/3 I believe. Or could be 1/3

Step-by-step explanation:

Because it is 3 cards and the probability is 2/3 for two 8ths

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What is the equation of a line with a slope of -2 that passes through the point (6, 8)?

iren [92.7K]

Answer:

2x + y = 20

Step-by-step explanation:

gradient = -2

x = 6, y = 8

y - 8 = -2(x - 6)

y - 8 = -2x + 12

2x + y = 12 + 8

2x + y = 20

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3 years ago

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-BARSIC- [3]

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3 years ago

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Please help with this calculus problem!

ELEN [110]

$\dfrac{\mathrm dQ}{\mathrm dt}=\dfrac{a(1-5bt^2)}{(1+bt^2)^4}$
$Q(t)=\displaystyle\int\frac{a(1-5bt^2)}{(1+bt^2)^4}\,\mathrm dt$

Let $t=\dfrac1{\sqrt b}\tan u$ , so that $\mathrm dt=\dfrac1{\sqrt b}\sec^2u\,\mathrm du$ . Then the integral becomes

$\displaystyle\frac a{\sqrt b}\int\frac{1-5b\left(\frac1{\sqrt b}\tan u\right)^2}{\left(1+b\left(\frac1{\sqrtb}\tan u\right)^2\right)^4}\sec^2u\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int\frac{1-5\tan^2u}{(1+\tan^2u)^4}\sec^2u\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int\frac{1-5\tan^2u}{\sec^6u}\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int(\cos^6u-5\sin^2u\cos^4u)\,\mathrm du$
$=\displaystyle\frac a{\sqrt b}\int(6\cos^2u-5)\cos^4u\,\mathrm du$

One way to proceed from here is to use the power reduction formula for cosine:

$\cos^{2n}x=\left(\dfrac{1+\cos2x}2\right)^n$

You'll end up with

$=\displaystyle\frac a{16\sqrt b}\int(5\cos2u+8\cos4u+3\cos6u)\,\mathrm du$
$=\displaystyle\frac a{16\sqrt b}\left(\frac52\sin2u+2\sin4u+\frac12\sin6u\right)+C$
$=\dfrac a{32\sqrt b}(5\sin2u+4\sin4u+\sin6u)+C$
$Q(t)=\dfrac{at}{(1+bt^2)^3}+C$

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3 years ago

What is the sqrt of (45n^5)?

djyliett [7]

3n^2 sqrt 5n

hope that helped (:

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mel-nik [20]

i dont get it either sorry

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