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Dafna1 [17]
3 years ago
7

What is 1,200,354,226,320 divided by 666,222,346

Mathematics
2 answers:
aleksklad [387]3 years ago
6 0

1,200,354,226,320 divided by 666,222,346 is 1801.73216

irina1246 [14]3 years ago
4 0

Answer:

1,801.73216

Step-by-step explanation:

hope this helps!

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Which of the following is not a US customary unit of volume?
zubka84 [21]
A pound is a unit of weight, not volume.
Good luck!
7 0
3 years ago
Read 2 more answers
Could anybody help me (pic attached)​
RideAnS [48]

Answer: 4/3

Step-by-step explanation:

What are the corresponding sides?

4 and 3

8 and 6

9 and 6.75

This is very easy, just make a fraction

4/3 = 4/3

8/6 = 4/3

9/6.75 = 4/3

Hence the answer is 4/3

5 0
3 years ago
The state where you live has a 6% sales tax. If you purchase a car for $9,800, what will you pay, including tax? (this is a two-
mestny [16]

Answer: $10,388

Step-by-step explanation:

The first step is to calculate the tax for the car:

9,800*0.06=588

The second step is to add the price of the car and the corresponding tax:

9,800+588=10,388

So, you will pay $10,388.

5 0
3 years ago
Workers have been down the street installing two blinking red lights in front of the fire station. The two lights are always tur
daser333 [38]

Answer:

After 88 seconds, they will blink together.

Step-by-step explanation:

Time after which the first light blinks = 8 seconds

Time after which the second light blinks = 22 seconds

Initially, both the lights blink together.

To find:

The time at which they will blink together again.

Solution:

Let us have a look at the time after which each light will blink.

First light will blink after:

8 s, 16 s, 24 s, 32 s, 40 s, 48 s, 56 s, 64 s, 72 s, 80 s, <em>88 s</em>, 96 s, 104 s, ......

Second light will blink after:

22 s, 44 s, 66 s, <em>88 s</em>, 110 s, .....

It can be seen that common time is 88 second, at which the both will blink together.

So, the answer is <em>88 s.</em>

<em></em>

This answer can also be computed by taking the LCM (Least Common Multiple) of both the numbers i.e. 8 and 22.

Using the factorization method:

8=\underline{2}\times 2\times 2

22=\underline{2}\times 11

Common is 2.

Therefore the LCM will be:

2\times 2\times2\times 11 = \bold{88}

So, the answer is <em>88 s.</em>

5 0
3 years ago
How do you solve his with working
AlexFokin [52]
Check the picture below.

a)

so the perimeter will include "part" of the circumference of the green circle, and it will include "part" of the red encircled section, plus the endpoints where the pathway ends.

the endpoints, are just 2 meters long, as you can see 2+15+2 is 19, or the radius of the "outer radius".

let's find the circumference of the green circle, and then subtract the arc of that sector that's not part of the perimeter.

and then let's get the circumference of the red encircled section, and also subtract the arc of that sector, and then we add the endpoints and that's the perimeter.

\bf \begin{array}{cllll}&#10;\textit{circumference of a circle}\\\\ &#10;2\pi r&#10;\end{array}\qquad \qquad \qquad \qquad &#10;\begin{array}{cllll}&#10;\textit{arc's length}\\\\&#10;s=\cfrac{\theta r\pi }{180}&#10;\end{array}\\\\&#10;-------------------------------

\bf \stackrel{\stackrel{green~circle}{perimeter}}{2\pi(7.5) }~-~\stackrel{\stackrel{green~circle}{arc}}{\cfrac{(135)(7.5)\pi }{180}}~+&#10;\stackrel{\stackrel{red~section}{perimeter}}{2\pi(9.5) }~-~\stackrel{\stackrel{red~section}{arc}}{\cfrac{(135)(9.5)\pi }{180}}+\stackrel{endpoints}{2+2}&#10;\\\\\\&#10;15\pi -\cfrac{45\pi }{8}+19\pi -\cfrac{57\pi }{8}+4\implies \cfrac{85\pi }{4}+4\quad \approx \quad 70.7588438888



b)

we do about the same here as well, we get the full area of the red encircled area, and then subtract the sector with 135°, and then subtract the sector of the green circle that is 360° - 135°, or 225°, the part that wasn't included in the previous subtraction.


\bf \begin{array}{cllll}&#10;\textit{area of a circle}\\\\ &#10;\pi r^2&#10;\end{array}\qquad \qquad \qquad \qquad &#10;\begin{array}{cllll}&#10;\textit{area of a sector of a circle}\\\\&#10;s=\cfrac{\theta r^2\pi }{360}&#10;\end{array}\\\\&#10;-------------------------------

\bf \stackrel{\stackrel{red~section}{area}}{\pi(9.5^2) }~-~\stackrel{\stackrel{red~section}{sector}}{\cfrac{(135)(9.5^2)\pi }{360}}-\stackrel{\stackrel{green~circle}{sector}}{\cfrac{(225)(7.5^2)\pi }{360}}&#10;\\\\\\&#10;90.25\pi -\cfrac{1083\pi }{32}-\cfrac{1125\pi }{32}\implies \cfrac{85\pi }{4}\quad \approx\quad 66.75884

7 0
3 years ago
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