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3 years ago

The music festival took 77 days to plan.

2 answers:

8 0

77 days turned into weeks means that you have to divide by seven.
77/7 = 11.
It took eleven weeks to plan (Answer choice D)

ivann1987 [24]3 years ago

7 0

Answer:

<em>(D)</em> is the answer

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The temperature in Antarctica during the winter can average -42.3 degrees Celsius. During the summer it can average around 11.52

Sindrei [870]

Answer:

-30.78!

Step-by-step explanation:

3 0

3 years ago

Q. The probability that a student passes a statistics test is 2/3 and the

kenny6666 [7]

Answer:

4/9 or 44.44%

Step-by-step explanation:

We have the following probabilities:

Probability of winning the statistical test = P (s) = 2/3

Probability of winning the mathematics and statistics test = P (s n m) = 14/45

Probability of winning at least one test = P (s u m) = 4/5

We have that, the probability of winning the mathematics test P (m) can be found in the following way:

P (s n m) = P (s) + P (m) - P (s u m)

replacing we have:

14/45 = 2/3 + P (m) - 4/5

P (m) = 14/45 - 2/3 + 4/5

p (m) = 4/9 = 0.444

Which is the same as there is a 44.44% probability that I will win the math test

5 0

3 years ago

How many 34 cup servings of granola can be measured from a box containing six cups of granola?

Bingel [31]

9514 1404 393

Answer:

Step-by-step explanation:

The number of servings can be found by dividing the quantity available by the amount per serving.

(6 cups)/(3/4 cup/serving) = (6)(4/3) servings = 8 servings

5 0

3 years ago

Explain how to determine if the two expressions are equivalent using x = 6 and x = 2.

Nataly_w [17]

<span>Two algebraic expressions are said to be equivalent if their values obtained by substituting the values of the variables are same.</span>

5 0

3 years ago

Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7

Digiron [165]

$y=x^r$
$\implies r(r-1)x^r+6rx^r+4x^r=0$
$\implies r^2+5r+4=(r+1)(r+4)=0$
$\implies r=-1,r=-4$

so the characteristic solution is

$y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}$

As a guess for the particular solution, let's back up a bit. The reason the choice of $y=x^r$ works for the characteristic solution is that, in the background, we're employing the substitution $t=\ln x$ , so that $y(x)$ is getting replaced with a new function $z(t)$ . Differentiating yields

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dz}{\mathrm dt}$
$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2z}{\mathrm dt^2}-\dfrac{\mathrm dz}{\mathrm dt}\right)$

Now the ODE in terms of $t$ is linear with constant coefficients, since the coefficients $x^2$ and $x$ will cancel, resulting in the ODE

$\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t$

Of coursesin, the characteristic equation will be $r^2+6r+4=0$ , which leads to solutions $C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}$ , as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If $z_1,z_2$ are the solutions to the characteristic equation of the ODE in terms of $z$ , then we can find another of the form $z_p=u_1z_1+u_2z_2$ where

$u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$
$u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$

where $W(z_1,z_2)$ is the Wronskian of the two characteristic solutions. We have

$u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t$

$u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t$

$\implies z_p=u_1z_1+u_2z_2$
$\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t$

and recalling that $t=\ln x\iff e^t=x$ , we have

$\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x$

4 0

2 years ago