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Serggg [28]
3 years ago
13

Find the least common multiple of these two expressions. 4v^7 w^5 and 10v^2w^4y^8

Mathematics
1 answer:
8090 [49]3 years ago
7 0

Answer:

20v^7w^5y^8.

Step-by-step explanation:

The LCM of 4 and 10 is 20.

of v^7 and v^2 is v^7

of w^5 and w^4 is w^5

So the answer is 20v^7w^5y^8.

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Which expression is equivalent to 11 - (-3 5/8) ?
scoray [572]

11 - (-3 5/8)

-negative sign and -negative sign= +positive sign

11+3 5/8

answer:

D. 11 + 3 5/8

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3 years ago
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7nadin3 [17]
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Sixty-eight percent of adults in a certain country believe that life on other planets is plausible. You randomly select five adu
Pavlova-9 [17]

Answer:

Mean = 3.4

Variance = 1.088

Standard deviation = 1.0431

Step-by-step explanation:

P=0.68\\ \\N=5

The mean of a binomial distribution with parameters N (the number of trials) and p (the probability of success for each trial) is m=N\cdot p

Thus,

m=5\cdot 0.68=3.4

The variance of the binomial distribution is s^2=Np(1-p), where s^2​​ is the variance of the binomial distribution, so

s^2=5\cdot 0.68\cdot (1-0.68)=3.4\cdot 0.32=1.088

The standard deviation s is the square root of the variance s^2, so

s=\sqrt{1.088}\approx 1.0431

3 0
2 years ago
Superman the Escape in California was once the fastest roller coaster in the world. Riders fall 415 feet in 7 seconds. Speeds re
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6 0
3 years ago
Problem 4: Solve the initial value problem
pishuonlain [190]

Separate the variables:

y' = \dfrac{dy}{dx} = (y+1)(y-2) \implies \dfrac1{(y+1)(y-2)} \, dy = dx

Separate the left side into partial fractions. We want coefficients a and b such that

\dfrac1{(y+1)(y-2)} = \dfrac a{y+1} + \dfrac b{y-2}

\implies \dfrac1{(y+1)(y-2)} = \dfrac{a(y-2)+b(y+1)}{(y+1)(y-2)}

\implies 1 = a(y-2)+b(y+1)

\implies 1 = (a+b)y - 2a+b

\implies \begin{cases}a+b=0\\-2a+b=1\end{cases} \implies a = -\dfrac13 \text{ and } b = \dfrac13

So we have

\dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = dx

Integrating both sides yields

\displaystyle \int \dfrac13 \left(\dfrac1{y-2} - \dfrac1{y+1}\right) \, dy = \int dx

\dfrac13 \left(\ln|y-2| - \ln|y+1|\right) = x + C

\dfrac13 \ln\left|\dfrac{y-2}{y+1}\right| = x + C

\ln\left|\dfrac{y-2}{y+1}\right| = 3x + C

\dfrac{y-2}{y+1} = e^{3x + C}

\dfrac{y-2}{y+1} = Ce^{3x}

With the initial condition y(0) = 1, we find

\dfrac{1-2}{1+1} = Ce^{0} \implies C = -\dfrac12

so that the particular solution is

\boxed{\dfrac{y-2}{y+1} = -\dfrac12 e^{3x}}

It's not too hard to solve explicitly for y; notice that

\dfrac{y-2}{y+1} = \dfrac{(y+1)-3}{y+1} = 1-\dfrac3{y+1}

Then

1 - \dfrac3{y+1} = -\dfrac12 e^{3x}

\dfrac3{y+1} = 1 + \dfrac12 e^{3x}

\dfrac{y+1}3 = \dfrac1{1+\frac12 e^{3x}} = \dfrac2{2+e^{3x}}

y+1 = \dfrac6{2+e^{3x}}

y = \dfrac6{2+e^{3x}} - 1

\boxed{y = \dfrac{4-e^{3x}}{2+e^{3x}}}

7 0
2 years ago
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