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laila [671]
2 years ago
15

Commercial agents earn 5% of the cost of each product they sell. If an agent earns

Mathematics
2 answers:
kicyunya [14]2 years ago
5 0

Based on the information the cost of the product is $40,000.

Given:

Percentage earn=5% or 0.05

Amount earned=$2,000

Let x represent the cost of the product

Hence:

Formulate

.05x = $2000  

Divide both sides by .05

x=$2,000/.05

x=$40,000

Inconclusion  the cost of the product is $40,000.

Learn more about cost here:brainly.com/question/824281

Artyom0805 [142]2 years ago
4 0

Answer:

40,000$

Step-by-step explanation:

I'ma be honest here. I literally just copied from the person above me to get points.

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Mark up is 40% and the cost to store is 162 dollars what is the selling price
Vanyuwa [196]

Answer:

$64.8

Step-by-step explanation:

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6 0
3 years ago
What is f(x)=2x^2+5 multiply by g(x)=2x
Kryger [21]

Answer:

4x^3 +10x

Step-by-step explanation:

for f(x) × g(x) you should multiply their functions :

(2x^2+5)×(2x) = 4x^3 +10x

6 0
3 years ago
Calculate the length b to two decimal places.
Andrej [43]

Answer:

B. 21.64

Step-by-step explanation:        

We have been given a triangle and we are asked to find the length of AC (b).

We will use law of cosines to find the length of side AC.

c^{2}=a^{2}+b^{2}-2ab \text{ cos }\theta

Upon substituting our given values in the formula we will get,

(AC)^{2}=15^{2}+12^{2}-2\times 15\times 12 \text{ cos}(106)  

(AC)^{2}=225+144-360\text{ cos}(106)      

(AC)^{2}=369-360(-0.275637355817)      

(AC)^{2}=369+99.22944809412    

(AC)^{2}=468.22944809412

Upon taking square root of both sides of our equation we will be get,

AC=\sqrt{468.22944809412}

AC=21.6386101238993629\approx 21.64  

Therefore, the length of b is 21.64 and option B is the correct choice.

6 0
3 years ago
sorry again for asking this but can someone explain this and give me the answer. Been having a hard time learning this.
Rudik [331]

<em>Hey</em><em>!</em><em>!</em><em>!</em>

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<em>In</em><em> </em><em>ques</em><em>tion</em><em> </em><em>no</em><em>.</em><em>5</em><em> </em><em>we</em><em> </em><em>have</em><em> </em><em>to</em><em><u> </u></em><em><u>add</u></em><em><u> </u></em><em><u>the </u></em><em><u>equation</u></em><em><u>.</u></em>

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3 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
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