Allison is rolling her hula hoop on the playground. the radius of her hula hoop is 35 ext{ cm}35 cm. what is the distance the hula hoop rolls in 44 full rotations? round your answer to the nearest ext{cm}cm.

Mathematics, 27.06.2019 20:30 P4thing

Allison is rolling her hula hoop on the playground. the radius of her hula hoop is 35 \text{ cm}35 cm. what is the distance the hula hoop rolls in 44 full rotations? round your answer to the nearest \text{cm}cm.

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Answer from: laurenlol756

distance covered in 44 full rotation = 9671 cm

Step-by-step explanation:

It is given that radius of hula hoop is 35 cm

circumference of the hula hoop = $2\pi (radius)$

= $2\pi (35)$

= $70\pi$

distance covered in one rotation = circumference of the hula hoop

= $70 \pi$

distance covered in 44 full rotation = $44(70)\pi$

= $3080\pi$

we have $\pi =3.14$

distance covered in 44 full rotation = 3080(3.14)

= 9671.2 cm

distance covered in 44 full rotation = 9671 cm ( rounding to nearest cm)

Answer from: faizaanahmed4p7gn2z

Distance covered by hula hoop in 44 rotations = 9671 cm

Step-by-step explanation:

In the given question,

radius of the hula hoop (r) = 35 cm

and full rotation = 44

Now,

circumference of the hula hoop = 2 $\pi$ r

Then,

distance covered in one rotation = circumference of the hula hoop

distance covered in 44 rotations = 44 × circumference of the hula hoop

Distance = 44 × 2 $\pi$ r = 44 × 2 × 3.14 × 35 = 9671.2 cm

Distance covered by hula hoop in 44 rotations = 9671 cm

Answer from: Quest

step-by-step explanation:

what is the figure? attach a picture.

Answer from: Quest

look at the picture.

$efgh\sim abcd\to2: 3$

ef : ab = 2 : 3

gf : cb = 2 : 3

gh : cd = 2 : 3

he : da = 2 : 3

∠a ≅ ∠e

∠b ≅ ∠f

∠c ≅ ∠g

∠d ≅ ∠h

$6: 3x=2: 3\to\dfrac{6}{3x}=\dfrac{2}{3}\to\dfrac{2}{x}=\dfrac{2}{3}\to2x=(2)(3)\to x=3$

use pythagorean theorem:

$a^2=8^2+(2\cdot3)^2\\\\a^2=64+36\\\\a^2=100\to a=\sqrt{100}\to a=10\ cm$

the length of the abcd trapezoid sides:

|ab| = 3 cm

|bc| = 10 cm

|cd| = 3(3) cm = 9 cm

|da| = 8 cm

the perimeter of abcd = 3 + 10 + 9 + 8 = 30 cm

the ratio of perimeters of the tarapezoids is equal 2 : 3.

$p_{efgh}: p_{abcd}=2: 3\to\dfrac{p_{efgh}}{p_{abcd}}=\dfrac{2}{3}$

substitute:

$\dfrac{p_{efgh}}{30}=\dfrac{2}{3}$ cross multiply

$3p_{efgh}=(30)(2)$ divide both sides by 3

$p_{efgh}=20\ cm$

$p_{abcd}=30\ cm\\\\p_{efgh}=20\ cm$

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Allison is rolling her hula hoop on the playground. the radius of her hula hoop is 35 \text{ cm}35 c...