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Rom4ik [11]
3 years ago
5

What is the sum of the prime factors of 2014

Mathematics
1 answer:
goblinko [34]3 years ago
7 0

The sum of prime factors of 2014 is 74

<h3><u>Solution:</u></h3>

Given that to find sum of prime factors of 2014

Let us first find the prime factors of 2014

A prime number is a whole number greater than 1 whose only factors are 1 and itself

"Prime Factorization" is finding which prime numbers multiply together to make the original number.

<em><u>Prime factors of 2014:</u></em>

The Prime Factorization is:

2014 = 2 \times 19 \times 53

Thus the prime factors of 2014 are 2, 19, 53

<em><u>Let us now find the sum of prime factors of 2014</u></em>

sum of prime factors of 2014 = 2 + 19 + 53 = 74

Thus the sum of prime factors of 2014 is 74

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Answer: $660

Step-by-step explanation:

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3 years ago
What is the effect on the graph of the function f(x) = 2^x when f(x) is replaced with f(x − 3)?
mel-nik [20]

Graph of f(x-3) is compressed by a factor of  \frac{1}{8} horizontally of f(x).

<u>Step-by-step explanation:</u>

We have, the graph of f(x)= 2^{x} , on replacing f(x) by f(x-3) we get:

f(x-3)= 2^{x-3} = \frac{2^{x}}{2^{3}} = \frac{1}{8} 2^{x} = \frac{1}{8} f(x).Below shown are the images for graph of f(x) and f(x-3). Both are functions are exponential , and so having exponential graph but f(x-3) is compressed by a factor of  \frac{1}{8} horizontally . Domain and range of both functions are same i.e. F(x) & f(x-3) domain & range are same , just difference in graph : f(x-3) = \frac{1}{8} f(x).

4 0
3 years ago
solve this system of linear equations. Separate the X- and Y- values with a comma. -9x+2y=-16 19x+3y=41​
Serjik [45]

Answer:

(2, 1)

Step-by-step explanation:

The best way to do this to avoid tedious fractions is to use the addition method (sometimes called the elimination method).  We will work to eliminate one of the variables.  Since the y values are smaller, let's work to get rid of those.  That means we have to have a positive and a negative of the same number so they cancel each other out.  We have a 2y and a 3y.  The LCM of those numbers is 6, so we will multiply the first equation by a 3 and the second one by a 2.  BUT they have to cancel out, so one of those multipliers will have to be negative.  I made the 2 negative.  Multiplying in the 3 and the -2:

3(-9x + 2y = -16)--> -27x + 6y = -48

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Now you can see that the 6y and the -6y cancel each other out, leaving us to do the addition of what's left:

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Now we will go back to either one of the original equations and sub in a 2 for x to solve for y:

19(2) + 3y = 41 so

38 + 3y = 41 and

3y = 3.  Therefore,

y = 1

The solution set then is (2, 1)

6 0
3 years ago
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The lcd would be 10x^3y^3

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3 years ago
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