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3 years ago

An aquarium tank holds 475 liters of water. How much is this in gallons? Use the following conversion: 1 gallon is 3.8 liters.

1 answer:

3 0

$\bf \begin{array}{ccll}
liters&gallons\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
3.8&1\\
475&g
\end{array}\implies \cfrac{3.8}{475}=\cfrac{1}{g}\implies g=\cfrac{475\cdot 1}{3.8}$

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Please help me with the below question.

VMariaS [17]

By letting

$y = \displaystyle \sum_{n=0}^\infty c_n x^{n+r}$

we get derivatives

$y' = \displaystyle \sum_{n=0}^\infty (n+r) c_n x^{n+r-1}$

$y'' = \displaystyle \sum_{n=0}^\infty (n+r) (n+r-1) c_n x^{n+r-2}$

a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to

$5r(r-1) c_0 x^{r-1} + \displaystyle \sum_{n=1}^\infty \bigg( (n+r+1) c_n + (n + r + 1) (5n + 5r + 1) c_{n+1} \bigg) x^{n+r} = 0$

Examine the lowest degree term $\left(x^{r-1}\right)$ , which gives rise to the indicial equation,

$5r (r - 1) + r = 0 \implies 5r^2 - 4r = r (5r - 4) = 0$

with roots at r = 0 and r = 4/5.

b) The recurrence for the coefficients $c_k$ is

$(k+r+1) c_k + (k + r + 1) (5k + 5r + 1) c_{k+1} = 0 \implies c_{k+1} = -\dfrac{c_k}{5k+5r+1}$

so that with r = 4/5, the coefficients are governed by

$c_{k+1} = -\dfrac{c_k}{5k+5} \implies \boxed{g(k) = -\dfrac1{5k+5}}$

c) Starting with $c_0=1$ , we find

$c_1 = -\dfrac{c_0}5 = -\dfrac15$

$c_2 = -\dfrac{c_1}{10} = \dfrac1{50}$

so that the first three terms of the solution are

$\displaystyle \sum_{n=0}^2 c_n x^{n + 4/5} = \boxed{x^{4/5} - \dfrac15 x^{9/5} + \frac1{50} x^{13/5}}$

4 0

2 years ago

Please help with this question

natita [175]

Answer:

$12\sin{(2\theta)}\cos{(\theta)}$

Step-by-step explanation:

The appropriate trig identity is ...

$\sin{a}+\sin{b}=2\sin{\left(\dfrac{a+b}{2}\right)}\cos{\left(\dfrac{a-b}{2}\right)}$

Here, you have a scale factor of 6 and a=3θ, b=θ. Filling in these values gives ...

$6\sin{3\theta}+6\sin{\theta}=12\sin{\left(\dfrac{3\theta+\theta}{2}\right)}\cos{\left(\dfrac{3\theta-\theta}{2}\right)}\\\\=12\sin{(2\theta)}\cos{(\theta)}$

8 0

3 years ago

Please help. Look at screenshot

Citrus2011 [14]

Answer:

Increasing:

x < 0

x > 2

Decreasing:

0 < x < 2

-1 < x < 2

x > 2

x < -1

Step-by-step explanation:

Function is increasing when it is going up as we go rightward

Function is decreasing when it is going down as we go rightward

We can see that as we move up (from negative infinity) until x = 0, the function is increasing. Also, as we go right from x = 2 towards positive infinity, the function is going up (increasing).

So,

Increasing:

x < 0

x > 2

The function is going down, or decreasing, at the in-between points of increasing, that is from 0 to 2, so that would be:

Decreasing:

0 < x < 2

When we want where the function is greater than 0, we basically want the intervals at which the function is ABOVE the x-axis [ f(x) > 0 ].

Looking at the graph, it is

from -1 to 2 (x axis)

and 2 to positive infinity

We can write:

-1 < x < 2

x > 2

Now we want when the function is less than 0, that is basically saying when the function is BELOW the x-axis.

This will be the other intervals than the ones we mentioned above in part (b).

Looking at the graph, we see that the graph is below the x-axis when it is less than -1, so we can write:

x < -1

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3 years ago

Pls help me ASAP no scamming:-)

Yanka [14]

Hi this was a different question from normal but okay.

Answer:

There has been an high speed chase that lead to an injury of a young girl. Isaac midway had fled the seen at North West 22nd, around 2:56 pm. With the girl laying in the street with a broken leg and fractured ankle. Have any evidence call crime stoppers.

Step-by-step explanation:

This provides criminal, time and location and injury. Good luck.

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DochEvi [55]

The answer is b y=6 Use photonath if any more equations like this one

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