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3 years ago

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Min and Jessica collected 60 plastic bottles. They divided the plastic bottles evenly between them and sold the bottles to a rec

ycling center for 10 cents each. Let n represent the amount of money, in cents, that Min and Jessica each earned from the recycling center. Which equation could be used to find the amount of money Min and Jessica each earned? n = 60 × 2 × 10 n = 60 ÷ 2 ÷ 10 n = 60 ÷ 2 × 10 n = 60 × 2 × 10

1 answer:

saveliy_v [14]3 years ago

5 0

Step-by-step explanation:

60/2=30

30x10= three dollars

n=60/2 x 10n

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What is the volume of the right rectangular prism below?

Vesnalui [34]

Answer:

Volume of right rectangular prism = 3.75 in³

Step-by-step explanation:

Given:

First side of rectangular prism = 2 inches

Second side of rectangular prism = 1 $\frac{1}{2}$ = 1.5 inches

Third side of rectangular prism = 1 $\frac{1}{4}$ = 1.25 inches

Find:

Volume of right rectangular prism:

Computation:

$Volume\ of\ right\ rectangular\ prism=lbh\\\\Volume\ of\ right\ rectangular\ prism=(2)(1.5)(1.25)\\\\ Volume\ of\ right\ rectangular\ prism= 3.75 in^3$

7 0

3 years ago

Use polar coordinates to find the volume of the given solid. Inside both the cylinder x2 y2 = 1 and the ellipsoid 4x2 4y2 z2 = 6

Anton [14]

The Volume of the given solid using polar coordinate is: $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

V= $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

<h3>What is Volume of Solid in polar coordinates?</h3>

To find the volume in polar coordinates bounded above by a surface z=f(r,θ) over a region on the xy-plane, use a double integral in polar coordinates.

Consider the cylinder, $x^{2}+y^{2} =1$ and the ellipsoid, $4x^{2}+ 4y^{2} + z^{2} =64$

In polar coordinates, we know that

$x^{2}+y^{2} =r^{2}$

So, the ellipsoid gives

$4{(x^{2}+ y^{2)} + z^{2} =64$

4( $r^{2}$ ) + $z^{2}$ = 64

$z^{2}$ = 64- 4( $r^{2}$ )

z=± $\sqrt{64-4r^{2} }$

So, the volume of the solid is given by:

V= $\int\limits^{2\pi}_ 0 \int\limits^1_0{} \, [\sqrt{64-4r^{2} }- (-\sqrt{64-4r^{2} })] r dr d\theta$

= $2\int\limits^{2\pi}_ 0 \int\limits^1_0 \, r\sqrt{64-4r^{2} } r dr d\theta$

To solve the integral take, $64-4r^{2}$ = t

dt= -8rdr

rdr = $\frac{-1}{8} dt$

So, the integral $\int\ r\sqrt{64-4r^{2} } rdr$ become

= $\int\ \sqrt{t } \frac{-1}{8} dt$

= $\frac{-1}{12} t^{3/2}$

= $\frac{-1}{12} (64-4r^{2}) ^{3/2}$

so on applying the limit, the volume becomes

V= $2\int\limits^{2\pi}_ {0} \int\limits^1_0{} \, \frac{-1}{12} (64-4r^{2}) ^{3/2} d\theta$

= $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(64-4(1)^{2}) ^{3/2} \; -(64-4(2)^{0}) ^{3/2} ] d\theta$

V = $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

Since, further the integral isn't having any term of $\theta$ .

we will end here.

The Volume of the given solid using polar coordinate is: $\frac{-1}{6} \int\limits^{2\pi}_ {0} [(60) ^{3/2} \; -(64) ^{3/2} ] d\theta$

Learn more about Volume in polar coordinate here:

brainly.com/question/25172004

#SPJ4

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The four highest scores at a diving meet were 9.08, 9.1, 9.15, and 9.06 points. Which of 3 of these numbers were the lowest out

julia-pushkina [17]

From lowest to highest:
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3 years ago

Read 2 more answers

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Alborosie

Answer:

Step-by-step explanation:

Median is the dot near the middle of each box. Since median of team one is to the left of that of team 2,

Team 1 has a lesser median than Team 2.

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Team 1 has a bigger IQR than Team 2.

See diagram.

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3 years ago

Really need help in these two questions!!!! Thank you math experts!!!!

const2013 [10]

<h3><u>Answers</u><u>:</u></h3>

1. $b = 7$ ,

$a = 12$

OR

$b = - 1$ ,

$a = - 4$

2. $= 10$

See attached images for explanation.

<h3 /><h3>I hope they help.</h3>

3 0

2 years ago