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Dmitrij [34]
3 years ago
9

Dexter has a fish tank shaped like a rectangular prism. It has a volume of 480 in^3 . The height of the tank is 8 inches taller

than the width, and the length of the tank is 6 inches longer than the width. What are the dimensions of the fish tank? Please show work 99 points
Mathematics
1 answer:
ella [17]3 years ago
4 0

Answer:

the height=12; the length=10; the width=4.

Step-by-step explanation:

1) the formula of the described volume is length*width*height=480 (in.³).

2) according to the condition the height=8+width; the length=6+width.

3) if to substitute all the values into the formula, ⇒ (6+width)*(8+width)*width=480.

(width)³+14(width)²+48width-480=0.

4) if to resolve the equation, the width=4.

The length=6+4=10; the height=8+4=12 (in.).

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30.2 - 13.7=n

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To get how far he ran you have to subtract 30.2 from 13.7 therefore giving you  16.5 which will be the total distance he ran after lunch.

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Solve the following system using either linear combination or substitution. Round any answers to the hundredths place. Show all
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So if y=-5x-23 then we can plug that value in for y in x-10y=6, so it becomes

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3 0
2 years ago
Convert from rectangular to polar coordinates: note: choose rr and θθ such that rr is nonnegative and 0≤θ<2π0≤θ<2π (a)(9,0
DochEvi [55]

Answer:

Step-by-step explanation:

Convert rectangle (x , y) to polar coordinates ( r , θ)

x=r  \cos \theta, y= r \sin \theta

r=\sqrt{x^2+y^2} , \theta =tan^-^1 (\frac{y}{x} )

a) converts (9, 0) to polar coordinates  ( r , θ)

r=\sqrt{x^2+y^2} \\\\=\sqrt{9^2+0} \\\\=9

\theta= \tan^-^1 (\frac{0}{9} )\\\\=0

b) Convert (18,\frac{18}{\sqrt{3} } ) to polar coordinates ( r, θ)

r = \sqrt{18^2+(\frac{18}{\sqrt{3} })^2 } \\\\=\sqrt{324+108} \\\\=\sqrt{432}

\frac{x}{y} \theta = \tan^-^1(\frac{\frac{18}{\sqrt{3} } }{18} )\\\\= \tan ^-^1(\frac{1}{\sqrt{3} } )\\\\= \frac{\pi}{6}

c)  converts (-5, 5) to polar coordinates  ( r , θ)

r =\sqrt{(-5)^2+(5)^2} \\\\=\sqrt{50} \\\\=5\sqrt{2}

\theta=\tan^-^1(\frac{5}{-5} )\\\\= \tan^-^1(-1)\\\\=\frac{3\pi}{4}

d)  converts (-1, √3) to polar coordinates  ( r , θ)

r=\sqrt{(-1)^2+(\sqrt{3})^2 } \\\\= \sqrt{4} \\\\=2

\theta=\tan^-^1(\frac{\sqrt{3} }{-1} )\\\=\tan^-^1(-\sqrt{3} )\\\\=\frac{2\pi}{3}

= \frac{2\pi}{\sqrt{3} }

6 0
3 years ago
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