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likoan [24]
2 years ago
5

Plz help fast

Mathematics
1 answer:
xz_007 [3.2K]2 years ago
3 0

Answer:

the function y=5x+5

plug in the hours 4.5 for x

y=5(4.5)+5

=27.5kilometers

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Kent caught an 8 pound 3 ounce trout in Pike Lake. The season's record catch so far is 129 ounces. Did Kent's fish beat the reco
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Yes Kent Did beat the record. There are sixteen ounces in one pound. 16•8=128+3 sooooo..131
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A rectangle has a length at most two more than seven times the width. If the perimeter is less than or equal to 52 units, find t
ella [17]

7x + 2 \leqslant 52
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May I know this answer?
alexira [117]

Answer:

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3 years ago
A number of runners enter a race. The times that each runner finished is based on every hour. Five runners finished the race in
Aleksandr [31]

The histogram for the data is given at the end of this answer.

<h3>What is an histogram?</h3>

An histogram is a graph that shows the number of times each element of x was observed.

In this problem:

  • The elements of x are given by the time in hours.
  • The histogram shows the number of runners that finished the race in x hours.

Hence, the graph given at the end of this answer is the histogram.

You can learn more about histogram at brainly.com/question/2962546

3 0
2 years ago
A right circular cylinder is inscribed in a sphere with diameter 4cm as shown. If the cylinder is open at both ends, find the la
SOVA2 [1]

Answer:

8\pi\text{ square cm}

Step-by-step explanation:

Since, we know that,

The surface area of a cylinder having both ends in both sides,

S=2\pi rh

Where,

r = radius,

h = height,

Given,

Diameter of the sphere = 4 cm,

So, by using Pythagoras theorem,

4^2 = (2r)^2 + h^2   ( see in the below diagram ),

16 = 4r^2 + h^2

16 - 4r^2 = h^2

\implies h=\sqrt{16-4r^2}

Thus, the surface area of the cylinder,

S=2\pi r(\sqrt{16-4r^2})

Differentiating with respect to r,

\frac{dS}{dr}=2\pi(r\times \frac{1}{2\sqrt{16-4r^2}}\times -8r + \sqrt{16-4r^2})

=2\pi(\frac{-4r^2+16-4r^2}{\sqrt{16-4r^2}})

=2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})

Again differentiating with respect to r,

\frac{d^2S}{dt^2}=2\pi(\frac{\sqrt{16-4r^2}\times -16r + (-8r^2+16)\times \frac{1}{2\sqrt{16-4r^2}}\times -8r}{16-4r^2})

For maximum or minimum,

\frac{dS}{dt}=0

2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})=0

-8r^2 + 16 = 0

8r^2 = 16

r^2 = 2

\implies r = \sqrt{2}

Since, for r = √2,

\frac{d^2S}{dt^2}=negative

Hence, the surface area is maximum if r = √2,

And, maximum surface area,

S = 2\pi (\sqrt{2})(\sqrt{16-8})

=2\pi (\sqrt{2})(\sqrt{8})

=2\pi \sqrt{16}

=8\pi\text{ square cm}

4 0
2 years ago
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