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cestrela7 [59]
3 years ago
11

An aquarium measures 61 cm long, 30.5 cm wide, and 30.5 cm high. If the tank is only filled to 80% capacity with water, how much

water is needed for this tank?
Mathematics
1 answer:
Oksanka [162]3 years ago
7 0
11349.05 (and whatever unit of measurement it would be)

The volume is 56745.25, and 80% of that would be 45396.2. In order to find the 20%, or how much water you need, you would subtract 45396.2 from 56745.25 and that leaves you with 11349.05.
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How can you get the variable alone in the<br> equation 25?<br> 5<br> a
WARRIOR [948]

Answer:

5

Step-by-step explanation:

5x5=25

or

5xa=25

a=5

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3 years ago
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Function or not a function?
n200080 [17]

Answer:

Not a function.

Step-by-step explanation:

There must only be one output per input but here we see the input 18 has two outputs ; 3 and 15 therefore this is not a function.

8 0
3 years ago
Subtract 17 from x which equals
Jet001 [13]
The answer to this problem is x - 17.
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working together Alberto and Amanda can paint a fence in 4.12 hours. had she done it alone it would have taken Amanda 10 hours.
nata0808 [166]

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Step-by-step explanation:

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3 years ago
Evaluate the following integral using trigonometric substitution
serg [7]

Answer:

The result of the integral is:

\arcsin{(\frac{x}{3})} + C

Step-by-step explanation:

We are given the following integral:

\int \frac{dx}{\sqrt{9-x^2}}

Trigonometric substitution:

We have the term in the following format: a^2 - x^2, in which a = 3.

In this case, the substitution is given by:

x = a\sin{\theta}

So

dx = a\cos{\theta}d\theta

In this question:

a = 3

x = 3\sin{\theta}

dx = 3\cos{\theta}d\theta

So

\int \frac{3\cos{\theta}d\theta}{\sqrt{9-(3\sin{\theta})^2}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9 - 9\sin^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{\theta})}}

We have the following trigonometric identity:

\sin^{2}{\theta} + \cos^{2}{\theta} = 1

So

1 - \sin^{2}{\theta} = \cos^{2}{\theta}

Replacing into the integral:

\int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{2}{\theta})}} = \int{\frac{3\cos{\theta}d\theta}{\sqrt{9\cos^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{3\cos{\theta}} = \int d\theta = \theta + C

Coming back to x:

We have that:

x = 3\sin{\theta}

So

\sin{\theta} = \frac{x}{3}

Applying the arcsine(inverse sine) function to both sides, we get that:

\theta = \arcsin{(\frac{x}{3})}

The result of the integral is:

\arcsin{(\frac{x}{3})} + C

8 0
3 years ago
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