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postnew [5]
1 year ago
7

Charlie is trying to cover the shape shown with paper. He will NOT be covering the top and the bottom. How much paper will he ne

ed? (Use 3.14 for TT) Select one: O 141.3 square meters O 423.9 square meters O 282.6 square meters O 339.12 square meters

Mathematics
1 answer:
DENIUS [597]1 year ago
6 0

You need to determine how much paper you need to cover the lateral side of the cylinder shown in the picture. For this, you have to calculate the surface area of the cylinder, which you can do using the following formula:

A=2\pi rh

Where

A is the area

π is the number pi, for the calculations we usually use up to the first two decimal values of this number, 3.14

r is the radius

h is the height of the cylinder

The given cylinder has a height of h=15m and a diameter of d=6m

To calculate the lateral area you need to use the radius. The diameter is twice the radius, so to determine the radius of the cylinder you have to divide the diameter by 2

\begin{gathered} r=\frac{d}{2} \\ r=\frac{6}{2} \\ r=3 \end{gathered}

Now you can calculate the lateral area as follows:

\begin{gathered} A=2\pi rh \\ A=2\cdot3.14\cdot3\cdot15 \\ A=282.6m^2 \end{gathered}

Charlie will need 282.6 m² to cover the lateral side of the cylinder.

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If I have line segmants with the measurements 3in, 5 in and 8 in; can I make a triangle?​
Dimas [21]

Answer:

Equilateral Triangle

Side a = 1

Side b = 1

Side c = 1

Angle ∠A = 60° = 1.0472 rad = π/3

Angle ∠B = 60° = 1.0472 rad = π/3

Angle ∠C = 60° = 1.0472 rad = π/3

C=60°B=60°A=60°b=1a=1c=1

Area = 0.43301

Perimeter p = 3

Semiperimeter s = 1.5

Height ha = 0.86603

Height hb = 0.86603

Height hc = 0.86603

Median ma = 0.86603

Median mb = 0.86603

Median mc = 0.86603

Inradius r = 0.28868

Circumradius R = 0.57735

Vertex coordinates: A[0, 0] B[1, 0] C[0.5, 0.86603]

Centroid: [0.5, 0.28868]

Inscribed Circle Center: [0.5, 0.28868]

Circumscribed Circle Center: [0.5, 0.28868]

Step-by-step explanation:

6 0
3 years ago
X/5 = 15 (Solving Linear Equations)
tiny-mole [99]
\dfrac{x}{5}=15\\
x=75
3 0
3 years ago
Solve this system of linear equations. separate the x- and y- values with a coma. -12x=-75-11y
ycow [4]
Solve the following system:
{-12 x = -11 y - 75 | (equation 1)
{-5 x = -11 y - 89 | (equation 2)
Express the system in standard form:
{-(12 x) + 11 y = -75 | (equation 1)
{-(5 x) + 11 y = -89 | (equation 2)
Subtract 5/12 × (equation 1) from equation 2:
{-(12 x) + 11 y = -75 | (equation 1)
{0 x+(77 y)/12 = (-231)/4 | (equation 2)
Multiply equation 2 by 12/77:
{-(12 x) + 11 y = -75 | (equation 1)
{0 x+y = -9 | (equation 2)
Subtract 11 × (equation 2) from equation 1:
{-(12 x)+0 y = 24 | (equation 1)
{0 x+y = -9 | (equation 2)

Divide equation 1 by -12:
{x+0 y = -2 | (equation 1)
{0 x+y = -9 | (equation 2)
Collect results:
Answer:  {x = -2                {y = -9

please note that the parentheses "{" should span over both equations but the editor doesn't allow that so I should it on both line, See attached example.










4 0
3 years ago
It takes Kevin 15 minutes to drive 52 miles.
Kaylis [27]

Answer:

C

Step-by-step explanation:

5 0
3 years ago
What is the solution set of x2 + 5x + 1 = 0?
DerKrebs [107]

Answer:

\huge\boxed{\left\{-\dfrac{5+\sqrt{21}}{2};\ \dfrac{-5+\sqrt{21}}{2}\right\}}

Step-by-step explanation:

x^2+5x+1=0\\\\\text{Use the quadratic formula:}\\\\\text{For}\ ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\\text{if}\ \Delta < 0,\ \text{then the equation has no real solution}\\\text{if}\ \Delta=0,\ \text{then the equation has one real solution}\ x=\dfrac{-b}{2a}\\\text{if}\ \Delta>0,\ \text{then the equation has two real solution}\ x=\dfrac{-b\pm\sqrt{\Delta}}{2a}

\text{We have}\\\\a=1,\ b=5,\ c=1\\\\\Delta=5^2-4(1)(1)=25-4=21>0\\\\x=\dfrac{-5\pm\sqrt{21}}{2(1)}=\dfrac{-5\pm\sqrt{21}}{2}

3 0
3 years ago
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