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3 years ago

Find the distance between the two points in simplest radical form. (4, 4) { and } (8, -3)

Mathematics

1 answer:

Temka [501]3 years ago

5 0

Answer:

привет, я не знаю, как ответить на вопрос

Step-by-step explanation:

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PLss answer I will give brainliest

marshall27 [118]

Answer:

I pretty sure it's 3/14

Step-by-step explanation:

There are 6 small fish and 8 large fish- find the total. 6+8=14 (14 will be the denominator)

Half of the small fish are red so 6/2=3 (3 is the numerator)

The probability of you picking one small red fish is 3/14

7 0

3 years ago

SOMEONE HELPPPPPPPPPPPPPPP!!!!

ziro4ka [17]

DE=2KL
8x+12=2(10x-9)
8x+12=20x-18
30=12x
x= 1 1/2
8x+12=DE
8(1.5)+12
12+12=DE
24=DE

Therefore, the measurement of DE is 24.

4 0

3 years ago

A certain library assesses fines for overdue books as follows. On the first day that a book is overdue, the total fine is $0.10.

HACTEHA [7]

Answer:

Option B. $0.70

Step-by-step explanation:

First day fine = $0.10

Second day fine (doubled or $0.30 which is lesser)

= $0.10 + $0.10 [$0.10 is lesser than $0.30]

= $0.20

Third day fine = $0.20 + $0.20 [ $0.20 is lesser than $0.30]

= $0.40

Fourth day fine = $0.40 + $0.30 = $0.70

[$0.30 is cheaper than doubling the amount $0.40 + $0.40]

Therefore, the total fine for a book on the fourth day is $0.70

4 0

3 years ago

SOLVE. integration of (1+v^2) /(1-v^3)

s2008m [1.1K]

$\displaystyle\int\frac{1+v^2}{1-v^3}\,\mathrm dv$

$1-v^3=(1-v)(1+v+v^2)$
$\implies\dfrac{1+v^2}{1-v^3}=\dfrac a{1-v}+\dfrac{b_0+b_1v}{1+v+v^2}$
$\implies\dfrac{1+v^2}{1-v^3}=\dfrac{a(1+v+v^2)+(b_0+b_1v)(1-v)}{1-v^3}$
$\implies 1+v^2=(a+b_0)+(a-b_0+b_1)v+(a-b_1)v^2$
$\implies\begin{cases}a+b_0=1\\a-b_0+b_1=0\\a-b_1=1\end{cases}\implies a=\dfrac23,b_0=\dfrac13,b_1=-\dfrac13$

So,

$\displaystyle\int\frac{1+v^2}{1-v^3}\,\mathrm dv=\dfrac23\int\frac{\mathrm dv}{1-v}+\dfrac13\int\frac{1-v}{1+v+v^2}\,\mathrm dv$

The first integral is easy. For the second, since the derivative of the denominator is $(1+v+v^2)=1+2v$ , we can add and subtract $3v$ to get

$\dfrac{1-v}{1+v+v^2}=\dfrac{1+2v-3v}{1+v+v^2}=\dfrac{1+2v}{1+v+v^2}-\dfrac{3v}{1+v+v^2}$

and for the first term employ a substitution. For the remaining term, we can complete the square in the denominator, then use a trigonometric substitution:

$\displaystyle\int\frac{1+2v}{1+v+v^2}\,\mathrm dv=\int\frac{\mathrm dt}t$

where $t=1+v+v^2\implies\mathrm dt=(1+2v)\,\mathrm dv$ , and

$\displaystyle\int\frac{3v}{1+v+v^2}\,\mathrm dv=3\int\frac v{\left(v+\frac12\right)^2+\frac34}\,\mathrm dv$

Then taking $v+\dfrac12=\dfrac{\sqrt3}2\tan s\implies \mathrm dv=\dfrac{\sqrt3}2\sec^2s\,\mathrm ds$ gives

$\displaystyle3\int\frac v{\left(v+\frac12\right)^2+\frac34}\,\mathrm dv=3\int\frac{\frac{\sqrt3}2\tan s-\frac12}{\left(\frac{\sqrt3}2\tan s\right)^2+\frac34}\left(\frac{\sqrt3}2\sec^2s\right)\,\mathrm ds$
$=\displaystyle\sqrt3\int(\sqrt3\tan s-1)\,\mathrm ds$
$=\displaystyle3\int\tan s\,\mathrm ds-\sqrt3\int\mathrm ds$

Now we're ready to wrap up.

$\displaystyle\int\frac{1+v^2}{1-v^3}\,\mathrm dv=\dfrac23\int\frac{\mathrm dv}{1-v}+\dfrac13\int\frac{1-v}{1+v+v^2}\,\mathrm dv$
$=\displaystyle-\frac23\ln|1-v|+\frac13\left(\int\frac{1+2v}{1+v+v^2}\,\mathrm dv-\int\frac{3v}{1+v+v^2}\,\mathrm dv\right)$
$=\displaystyle-\frac23\ln|1-v|+\frac13\int\frac{\mathrm dt}t-\frac13\left(3\int\tan s\,\mathrm ds-\sqrt3\int\mathrm ds\right)$
$=\displaystyle-\frac23\ln|1-v|+\frac13\ln|t|-\int\tan s\,\mathrm ds+\frac1{\sqrt3}\int\mathrm ds$
$=\displaystyle-\frac23\ln|1-v|+\frac13\ln|1+v+v^2|+\ln|\cos s|+\frac s{\sqrt3}+C$
$=\displaystyle-\frac23\ln|1-v|+\frac13\ln|1+v+v^2|+\ln\left|\frac{\sqrt3}{2\sqrt{1+v+v^2}}\right|+\frac1{\sqrt3}\tan^{-1}\frac{2v+1}{\sqrt3}+C$

This can be simplified a bit using some properties of logarithms to obtain

$=\displaystyle-\frac23\ln|1-v|+\frac13\ln(1+v+v^2)+\left(\ln\frac{\sqrt3}2-\frac12\ln(1+v+v^2)\right)+\frac1{\sqrt3}\tan^{-1}\frac{2v+1}{\sqrt3}+C$
$=\displaystyle-\frac23\ln|1-v|-\frac16\ln(1+v+v^2)+\frac1{\sqrt3}\tan^{-1}\frac{2v+1}{\sqrt3}+C$

6 0

3 years ago

Anika buys a pair of shin guards for $8.58 and serveral pairs of soccer socks for $4.29 per pair. If she spends 25.74 before tax

Lena [83]

Answer:

4 socks

Step-by-step explanation:

4 0

4 years ago