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3 years ago

Help me solve this problem please

Mathematics

1 answer:

saul85 [17]3 years ago

5 0

Answer:

x=0

Step-by-step explanation:

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What is the center and radius of the circle? *<br> (x - 2)² + (y – 9)^2= 4

Rudiy27

the center is (2,9)

and the radius is 2

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3 years ago

The blueprint for a new swimming pool uses the scale 3/4 inch = 2 feet. If the actual dimensions of the pool will be 8 ft wide b

weeeeeb [17]

Answer:

The dimensions of the pool on the blueprint would be 3 in wide by 4.5 in long.

Step-by-step explanation:

First I turned 3/4 into a decimal, which would be 0.75. The I divided the dimensions of the pool by two. This way I could multiply the .75 by how many sets of two feet there were in the dimensions. Finally I multiplied 4 by .75 and I multiplied 6 by .75. That is how I found my answer.

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3 years ago

The x-coordinate of the point (50, 55) is ___________.

marysya [2.9K]

The x-coordinate of the point (50, 55) is 50

<h3>What is graph?</h3>

A graph can be defined as a pictorial representation or a diagram that represents data or values.

The distance of point (50, 55) perpendicular to y-axis

Thus, perpendicular distance = 50 units

The x-coordinate of the point (50, 55) is 50

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2 years ago

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. A right cylindrical drum is to hold 7.35 cubic feet of liquid. Find the dimensions (radius of the base and height) of the drum

Drupady [299]

Answer:

$r=1.05\ \text{ft}$

$h=2.12\ \text{ft}$

$20.91\ \text{ft}^3$

Step-by-step explanation:

r = Radius

h = Height

Volume of cylinder = $7.35\ \text{ft}^3$

$V=\pi r^2h\\\Rightarrow h=\dfrac{V}{\pi r^2}\\\Rightarrow h=\dfrac{7.35}{\pi r^2}$

Surface area is given by

$A=2\pi r^2+2\pi rh\\\Rightarrow A=2\pi r^2+2\pi r\dfrac{7.35}{\pi r^2}\\\Rightarrow A=2\pi r^2+\dfrac{14.7}{r}$

Differentiating with respect to radius we get

$\dfrac{dA}{dr}=4\pi r-\dfrac{14.7}{r^2}$

Equating with zero we get

$0=4\pi r-\dfrac{14.7}{r^2}\\\Rightarrow 4\pi r=\dfrac{14.7}{r^2}\\\Rightarrow r^3=\dfrac{14.7}{4\pi}\\\Rightarrow r=(\dfrac{14.7}{4\pi})^{\dfrac{1}{3}}\\\Rightarrow r=1.05$

$\dfrac{d^2A}{dr^2}=4\pi-2\times \dfrac{14.7}{r^3}\\\Rightarrow \dfrac{d^2A}{dr^2}=4\pi+2\times \dfrac{14.7}{1.05^3}\\\Rightarrow \dfrac{d^2A}{dr^2}=37.96>0$