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sergey [27]
3 years ago
13

What is 4855/200 simplified

Mathematics
1 answer:
EleoNora [17]3 years ago
3 0

Answer:

The answer is 2427.5

Step-by-step explanation:

The fraction consists of two numbers and a fraction bar: 4,855/200

The number above the bar is called numerator: 4,855

The number below the bar is called denominator: 200

The fraction bar means that the two numbers are dividing themselves.

To get fraction's value divide the numerator by the denominator:

Value = 4,855 ÷ 200

To calculate the greatest common factor, GCF:

1. Build the prime factorizations of the numerator and denominator.

2. Multiply all the common prime factors, by the lowest exponents.

Factor both the numerator and denominator, break them down to prime factors:

Prime Factorization of a number: finding the prime numbers that multiply together to make that number.

4,855 = 5 × 971;

4,855 is a composite number;

In exponential notation:

200 = 2 × 2 × 2 × 5 × 5 = 23 × 52;

200 is a composite number;

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Answer:

f(2)=0. The third option is correct.

Step-by-step explanation:

Suppose we have a polynomial called f(x) and we know (x-2) is a factor of f. This means that we can express f(x) as:

f(x)=(x-2)\cdot g(x)

Being g(x) another unknown function.

Option 1: f(0)=(0-2)\cdot g(0)=-2\cdot g(0). Since we don't know the value of g(0), we cannot assure f(0)=-2. Incorrect

Option 2: f(-2)=(-2-2)\cdot g(-2)=-4\cdot g(-2). Since we don't know the value of g(-2), we cannot assure f(-2)=0. Incorrect

Option 3: f(2)=(2-2)\cdot g(2)=0\cdot g(0)=0. Regardless of the value of g(0), f(0) must be 0. Correct

Option 4: f(0)=(0-2)\cdot g(0)=-2\cdot g(0). Since we don't know the value of g(0), we cannot assure f(0)=2. Incorrect

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3 years ago
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Using Pythagorean Theorem, 25+64=sq root of 89.
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35 lbs of apples cost $420. <br> How much would 6 lbs cost?
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Step-by-step explanation:

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For what values of x is f(x) = |x + 1| differentiable? I'm struggling my butt off for this course
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By definition of absolute value, you have

f(x) = |x+1| = \begin{cases}x+1&\text{if }x+1\ge0 \\ -(x+1)&\text{if }x+1

or more simply,

f(x) = \begin{cases}x+1&\text{if }x\ge-1\\-x-1&\text{if }x

On their own, each piece is differentiable over their respective domains, except at the point where they split off.

For <em>x</em> > -1, we have

(<em>x</em> + 1)<em>'</em> = 1

while for <em>x</em> < -1,

(-<em>x</em> - 1)<em>'</em> = -1

More concisely,

f'(x) = \begin{cases}1&\text{if }x>-1\\-1&\text{if }x

Note the strict inequalities in the definition of <em>f '(x)</em>.

In order for <em>f(x)</em> to be differentiable at <em>x</em> = -1, the derivative <em>f '(x)</em> must be continuous at <em>x</em> = -1. But this is not the case, because the limits from either side of <em>x</em> = -1 for the derivative do not match:

\displaystyle \lim_{x\to-1^-}f'(x) = \lim_{x\to-1}(-1) = -1

\displaystyle \lim_{x\to-1^+}f'(x) = \lim_{x\to-1}1 = 1

All this to say that <em>f(x)</em> is differentiable everywhere on its domain, <em>except</em> at the point <em>x</em> = -1.

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3 years ago
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