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3 years ago

The surface area of a cylinder?

Mathematics

2 answers:

goldfiish [28.3K]3 years ago

8 0

Answer:

18.85 $ft^2\\$

*You should run the numbers yourself as well. Sometimes different calculators will get marginally different numbers or use a different rounding for $\pi$ that gives a slightly different answer*

Step-by-step explanation:

Surface area of a cylinder: $2\pi rh+2\pi r^2$

Where h is the height and r is the radius. Remember that the radius is half the diameter, and the diameter is a straight line that passes through a circle.

I could be wrong, but I think you had the correct equation but used the diameter in stead of the radius to get 50.36.

Radius: 1 Height: 2

Plug numbers into equation:

$A=2\pi (1)(2)+2\pi (1)^2= 18.8495. . .$

I hope that helps!

ioda3 years ago

7 0

Answer:

18. 84 ft² or 18.85 ft² when rounded to the nearest tenth

Step-by-step explanation:

2πrh+2πr²

2× 3.14 × 1 × 2= 12.56

2 × 3.14 × 1² = 6.28

12.56 + 6.28 = 18.84

Have a great day :)

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How does adding the log together automatically mean that it is a factorial?

andreyandreev [35.5K]

Answer: The answer is given below.

Step-by-step explanation: We are given an equality involving logarithm and we are to show the implication of L.H.S. to R.H.S.

We will be using the following two properties of logarithm:

$(i)~\log_ba=\dfrac{1}{\log_ab},\\\\\\(ii)~log_ab+\log_ac=\log_a(bc).$

The proof is as follows:

$L.H.S.\\\\\\=\dfrac{1}{\log_2N}+\dfrac{1}{\log_3N}+\dfrac{1}{\log_4N}+\cdots+\dfrac{1}{\log_{100}N}\\\\\\=\log_N2+\logN3+\log_N4+\cdots+\log_N100\\\\=\log_N\{2.3.4...100\}\\\\=\log_N\{1.2.3.4...100\}\\\\=\log_N{100!}\\\\=\dfrac{1}{\log_{100!}N}\\\\=R.H.S.$

Hence proved.

6 0

3 years ago

A circle is centered at J(3, 3) and has a radius of 12.

stealth61 [152]

Answer:

$(-6,\, -5)$ is outside the circle of radius of $12$ centered at $(3,\, 3)$ .

Step-by-step explanation:

Let $J$ and $r$ denote the center and the radius of this circle, respectively. Let $F$ be a point in the plane.

Let $d(J,\, F)$ denote the Euclidean distance between point $J$ and point $F$ .

In other words, if $J$ is at $(x_j,\, y_j)$ while $F$ is at $(x_f,\, y_f)$ , then $\displaystyle d(J,\, F) = \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}$ .

Point $F$ would be inside this circle if $d(J,\, F) < r$ . (In other words, the distance between $F\!$ and the center of this circle is smaller than the radius of this circle.)

Point $F$ would be on this circle if $d(J,\, F) = r$ . (In other words, the distance between $F\!$ and the center of this circle is exactly equal to the radius of this circle.)

Point $F$ would be outside this circle if $d(J,\, F) > r$ . (In other words, the distance between $F\!$ and the center of this circle exceeds the radius of this circle.)

Calculate the actual distance between $J$ and $F$ :

$\begin{aligned}d(J,\, F) &= \sqrt{(x_j - x_f)^{2} + (y_j - y_f)^{2}}\\ &= \sqrt{(3 - (-6))^{2} + (3 - (-5))^{2}} \\ &= \sqrt{145} \end{aligned}$ .

On the other hand, notice that the radius of this circle, $r = 12 = \sqrt{144}$ , is smaller than $d(J,\, F)$ . Therefore, point $F$ would be outside this circle.