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12345 [234]
3 years ago
13

A team of scientists is traveling by boat to study a pod, or group, of dolphins. The equation y = 0.2x can be used to find the d

istance, y, in miles the boat travels in x minutes. The dolphins are currently 8.4 mi from the boat. If the dolphins stay to where they are, how long will it take the boat to reach them? Show your work.
Mathematics
1 answer:
Alenkinab [10]3 years ago
6 0

Answer:

42 minutes.

Step-by-step explanation:

We know that:

y = 0.2*x

is the equation that that defines the distance, in miles, that the boat travels in x minutes.

We know that the dolphins are at 8.4 miles from the boat.

Then when we have y = 8.4, this will mean that the boat reached the dolphins.

Then we need to solve the equation:

y = 8.4 = 0.2*x

for x

To do that, we just need to isolate x in one side of the equation, so we can just divide both sides by 0.2 to get:

8.4/0.2 = (0.2*x)/0.2

42 = x

And x is time in minutes, thus, the boat needs to travel for 42 minutes to reach the dolphins.

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Cameron has 2 video game holder stands each can hold 2 rows of 20 games write and evaluate a numerical expression to find the to
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Answer:

so this would be the equation

2(2*10)=v

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This shows the number of games total in both holders.


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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

3 0
2 years ago
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