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andrew11 [14]
2 years ago
14

Please help asap thank you!!

Mathematics
1 answer:
svet-max [94.6K]2 years ago
6 0

Answer:

No

yes

No

Step-by-step explanation:

yes it is a scalene triangle

the area of the triangle is 28.5

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What is the surface area of the cylinder with height 7 cm and radius 5 cm? Round
nalin [4]

Answer:

376.991 cm²

Step-by-step explanation:

SA_{cylinder} = 2\pi r^{2}  + 2\pi rh

Surface Area = 2π(5²) + 2π(5)(7)

                      = 50π + 70π

                      = 120π

                      = 376.9911184... ≈ 376.991

<h2>Derive Surface Area of A Cylinder</h2>

If you take a look at a net of a cylinder, you can see it is composed of two circles(bases) and a rectangular strip with the length of the diameter of the circle

The formula for the area of a circle = πr², and since there are 2, it is 2πr²

The formula for circumference of a circle = 2πr, and since we are multiplying that by the height by the height of the cylinder, it is 2πrh

∴ 2πr² + 2πrh

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UNLI Quiz Active 1 2 3 5 11 TIME What should be done to solve the equation? X+ 14 = 21 Add 14 to both sides of the equation. Sub
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6) the circular base of a hemisphere of radius 2 rests on the base of a square pyramid of height 6. the hemisphere is tangent to
Rasek [7]

The length of the square base is thus 2 x 3\sqrt{2}/2 = 3\sqrt{2} = A

<h3>What is hemisphere?</h3>

Consequently, a hemisphere is a 3D geometric object that is made up of half of a sphere, with one side being flat and the other being a bowl-like shape. It is created by precisely cutting a spherical along its diameter, leaving behind two identical hemispheres.

EXPLANATION; Let ABCDE be the pyramid with ABCD as the square base. Let O and M be the center of square ABCD and the midpoint of side AB respectively. Lastly, let the hemisphere be tangent to the triangular face ABE at P.

Notice that triangle EOM has a right angle at O. Since the hemisphere is tangent to the triangular face ABE at P, angle EPO is also 90 degree. Hence, triangle EOM is similar to triangle EPO.

OM/2 = 6/EP

OM = 6/EP x 2

OM = 6\sqrt{6^2 - 2^2} x 2 = 3\sqrt{2}/2}

The length of the square base is thus 2 x 3\sqrt{2}/2 = 3\sqrt{2} = A

To know more about Hemisphere, visit;

brainly.com/question/13625065?referrer=searchResults

#SPJ4

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