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3 years ago

What are the common multiples for 5 and 10? Please answer. :(

Mathematics

2 answers:

NeTakaya3 years ago

4 0

Answer:

50?? I think that's the answer

White raven [17]3 years ago

3 0

Answer:

1,5

Hope it helps

Have a good day

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Fill in the blank with the correct number.

kvasek [131]

Answer:D

Step-by-step explanation:

3 0

3 years ago

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At a school carnival, the diameter of the mat of a trampoline is 10 feet and the diameter of its metal frame is 12 feet. What is

Zielflug [23.3K]

Answer:

Step-by-step explanation:

circumference = pi * diameter

circumference of metal frame = pi * (diameter of metal frame)

(diameter of metal frame) = 12

Circumference of Metal Frame = 3.14 * 12 = 37.68

therefore D: 37.7 feet would be your answer

3 0

3 years ago

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Can somebody help me with this?

Natasha_Volkova [10]

Step-by-step explanation:

Hey there!

By looking through figure, l and m are parallel lines and a transversal line passes through the lines.

Now,

7x + 12° = 12x - 28°( alternate angles are equal)

12°+28° = 12x - 7x

40° = 5x

$x = \frac{40}{5}$

Therefore, x = 8°

Now,

12x - 28° + 9y - 77 = 180° ( being linear pair)

12×8° - 28° + 9y -77° = 180°

96° - 28° + 9y - 77° = 180°

-9 + 9y = 180°

9y = 180° + 9°

y = 189°/9

Therefore, y = 21°

Therefore, X = 8° and y= 21° .

Hope it helps....

4 0

4 years ago

How to do 314,207 in standard form??

olya-2409 [2.1K]

314 207 = 300 000 + 10 000 + 4 000 + 200 + 7

4 0

4 years ago

Find (a) the arc length and (b) the area of a sector.

Brut [27]

Answer:

a) 23.56 ft (2 dp)

b) 58.90 ft² (2 dp)

Step-by-step explanation:

Formula

$\textsf{Arc length}=r \theta$

$\textsf{Area of a sector}=\dfrac{1}{2}r^2 \theta$

$\quad \textsf{(where r is the radius and}\:\theta\:{\textsf{is the angle in radians)}$

Calculation

Given:

$\theta=\dfrac{3 \pi}{2}$
r = 5 ft

$\begin{aligned}\implies \textsf{Arc length} & =r \theta\\& = 5\left(\dfrac{3 \pi}{2}\right)\\& = \dfrac{15}{2} \pi \\& = 23.56\: \sf ft\:(2\:dp)\end{aligned}$

$\begin{aligned} \implies \textsf{Area of a sector}& =\dfrac{1}{2}r^2 \theta\\\\ & = \dfrac{1}{2}(5^2) \left(\dfrac{3 \pi}{2}\right)\\\\& = \dfrac{25}{2}\left(\dfrac{3 \pi}{2}\right)\\\\ & = \dfrac{75}{4} \pi \\\\& = 58.90 \: \sf ft^2\:(2\:dp)\end{aligned}$

6 0

3 years ago