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VLD [36.1K]
3 years ago
14

How do you divide a fraction

Mathematics
2 answers:
mariarad [96]3 years ago
7 0
"Dividing<span> by </span>fractions<span> is just like multiplying </span>fractions<span>, except for one additional step. To </span>divide<span> any number by a </span>fraction<span>: First step: Find the reciprocal of the </span>fraction<span>. Second step: Multiply the number by the reciprocal of the </span>fraction<span>." -Google</span>
ella [17]3 years ago
3 0
You break it down into a whole number then you divide it.
You might be interested in
Acellus
taurus [48]

Answer:

The Probability is 63.7%

Step-by-step explanation:

We need to calculate the area of the circle and the area of the square

From the question, we can see that the square is of side 4 √2 cm

The area of the square is simply the square of its side length

We have this as 4√2 * 4√2 = 16 * 2 = 32 cm^2

The area of the circle can be calculated using the formula for the area of a circle, with the radius being 4 cm

So, we have the area of the circle as;

Pi * r^2

= 22/7 * 4 * 4 = 50.3 cm^2

so, to find the probability, we need to divide the area of the circle by the area of the square

we have this as;

32/50.3 = 0.6366

In percentage, we simply multiply by 100% to give

0.6366 * 100% = 63.66%

To the nearest tenth, this is 63.7%

8 0
3 years ago
How would I solve this?
lutik1710 [3]

Answer:

It is rotated by 72 degrees.

Step-by-step explanation:

  • Since it is a regular polygon,

        when u connect all the corners of it to the middle of the polygon, they       will meet at a point i.e, CENTER.

  • The sum of the angles subtended by all the sided at the center will be 360 degrees.
  • As there are 60 sides, the angle subtended by each side at the center will be 6 degrees.

Because,

\frac{360}{60} = 6

  • As the polygon rotates every minute and it is rotated for 12 minutes,

12*6 = 72

( For every minute, it will be rotated by 6 degrees.

so, for 12 minutes it should be rotated by 12 times 6 ( 12*6) = 72 degrees)

  • So, after 12 minutes it will be rotated by 72 degrees.
5 0
3 years ago
Let $s$ be a subset of $\{1, 2, 3, \dots, 100\}$, containing $50$ elements. how many such sets have the property that every pair
Tamiku [17]

Let A be the set {1, 2, 3, 4, 5, ...., 99, 100}.

The set of Odd numbers O = {1, 3, 5, 7, ...97, 99}, among these the odd primes are :

P={3, 5, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}

we can count that n(O)=50 and n(P)=24.

 

 

Any prime number has a common factor >1 with only multiples of itself.

For example 41 has a common multiple >1 with 41*2=82, 41*3=123, which is out of the list and so on...

For example consider the prime 13, it has common multiples >1 with 26, 39, 52, 65, 78, 91, and 104... which is out of the list.

Similarly, for the smallest odd prime, 3, we see that we are soon out of the list:

3, 3*2=6, 3*3=9, ......3*33=99, 3*34=102.. 

we cannot include any non-multiple of 3 in a list containing 3. We cannot include for example 5, as the greatest common factor of 3 and 5 is 1.

This means that none of the odd numbers can be contained in the described subsets.

 

 

Now consider the remaining 26 odd numbers:

{1, 9, 15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, 69, 75, 77, 81, 85, 87, 91, 93, 95, 99}

which can be written in terms of their prime factors as:

{1, 3*3, 3*5, 3*7, 5*5,3*3*3, 3*11,5*7, 3*13, 2*2*3*3, 7*7, 3*17, 5*11 , 3*19,3*21, 5*13, 3*23,3*5*5, 7*11, 3*3*3*3, 5*17, 3*29, 7*13, 3*31, 5*19, 3*3*11}

 

1 certainly cannot be in the sets, as its common factor with any of the other numbers is 1.

3*3 has 3 as its least factor (except 1), so numbers with common factors greater than 1, must be multiples of 3. We already tried and found out that there cannot be produced enough such numbers within the set { 1, 2, 3, ...}

 

3*5: numbers with common factors >1, with 3*5 must be 

either multiples of 3: 3, 3*2, 3*3, ...3*33 (32 of them)

either multiples of 5: 5, 5*2, ...5*20 (19 of them)

or of both : 15, 15*2, 15*3, 15*4, 15*5, 15*6 (6 of them)

 

we may ask "why not add the multiples of 3 and of 5", we have 32+19=51, which seems to work.

The reason is that some of these 32 and 19 are common, so we do not have 51, and more important, some of these numbers do not have a common factor >1:

for example: 3*33 and 5*20

so the largest number we can get is to count the multiples of the smallest factor, which is 3 in our case.

 

By this reasoning, it is clear that we cannot construct a set of 50 elements from {1, 2, 3, ....}  containing any of the above odd numbers, such that the common factor of any 2 elements of this set is >1.

 

What is left, is the very first (and only) obvious set: {2, 4, 6, 8, ...., 48, 50}

 

<span>Answer: only 1: the set {2, 4, 6, …100}</span>

8 0
3 years ago
Can someone please help me with this and show work as well?? thank you so much!
Schach [20]
Use a compass to make a circle around both lasers at the same degree and they should over lap. where they over lap use your straight edge to make a direct line from the point where your at through where they cross. that will be your answer, hope i helped.
8 0
3 years ago
Y = 2x - 3<br> x + 2y = 4
uranmaximum [27]

Answer:

x=2 , y = 1

Step-by-step explanation:

x + 2(2x-3) = 4

x+4x-6=4

5x-6=4

5x= 10

x=10/5

x=2

subs x=2 into x +2y=4

2+2y=4

2y=2

y=2/2

y=1

6 0
3 years ago
Read 2 more answers
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