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julia-pushkina [17]
3 years ago
13

Bob can drive around the racetrack 72 times for $12. How much would it cost for Bob to drive the race track 27 times?

Mathematics
1 answer:
Naddika [18.5K]3 years ago
8 0
$4.50. To find how many laps we can go for one dollar, in this case, we divide 72 by 12. If we do this, we know that we can go 6 laps for 1 dollar. 27 divided by 6 is equal to 4.5, therefore, it will cost $4.50. Pls mark me brainliest if this is correct :)
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5. A gardening club has 21 members, of which 13 are males and the rest are
Misha Larkins [42]

If you know that there are 21 members and 13 are females, if you subtract 13 from 21, you will get 8. There are 8 females in the gardening club.

Now you have to figure out the ratio to females to all club members. You already know that there are 8 females and 21 club members. The ratio would be 8:21.

I hope you found this helpful :)

3 0
3 years ago
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Solve the equation for W and L
Roman55 [17]

Answer:

Step-by-step explanation:

Cross multiply and get:

\sqrt{7}L = 5w

L = 5w/\sqrt{7}

W = \sqrt{7}L/5

4 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
Help with this pls?????
Alenkinab [10]

Answer:

83.74 in squared

Step-by-step explanation:

The area of the original rectangle is: 14 * 8 = 112 in squared.

The area of a circle is \pi r^{2}, where r is the radius. In this case, we know the diameter is 6, so the radius is 3. Then, the area of the circle is:

3.14 * 3^2 = 3.14 * 9 = 28.26 in squared.

The remaining area is just the total minus the circle area:

112 - 28.26 = 83.74 in squared

6 0
3 years ago
Read 2 more answers
The volume of a box is found by multiplying its length, width, and height. The three sides are 0.5 foot, 0.75 foot, and 0.4 foot
just olya [345]
(1/2)×(3/4)×(2/5)=5/40=1/8

as a decimal, this is .125
for scientific notation, you must move the decimal one place to the right, so the answer is

1.25×10^-1 ft^3
4 0
3 years ago
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