All categories

3 years ago

What is not a problem modular division is used for?

Mathematics

1 answer:

Oksana_A [137]3 years ago

4 0

Answer:

First let's define what modular arithmetic is, what would come is an arithmetic system for equivalence classes of whole numbers called congruence classes.

Now, the modular division is the division in modular arithmetic.

Answering the question, a modular division problem like ordinary arithmetic is not used, division by 0 is undefined. For example, 6/0 is not allowed. In modular arithmetic, not only 6/0 is not allowed, but 6/12 under module 6 is also not allowed. The reason is that 12 is congruent with 0 when the module is 6.

You might be interested in

If you are selling bag of candy would you want ti sell the candy in ounces or pounds? Explain.

7nadin3 [17]

You would want to sell the bag of candy in ounces, unless you have way too much candy.

8 0

3 years ago

Read 2 more answers

Need help on khan NOW! question is on picture. will report link and mark brainliest if right

Genrish500 [490]

Answer: R= 1

Explanation: no matter how you do it you still get 1 bc its x divided by y which is 1

3 0

2 years ago

Use base ten blocks to solve 53/4

Crazy boy [7]

if you have base ten blocks you can figure out the answer

6 0

3 years ago

Find the x and y intercepts if the function

Tresset [83]

I belive there is none because it lands on the origin for y.

7 0

3 years ago

Read 2 more answers

|1/2b-8|=|1/4b-1|<br> b=____ and ____

Deffense [45]

Answer:

b = 12 and 28

Step-by-step explanation:

The absolute value equation |1/2b-8| = |1/4b-1| resolves to a piecewise linear function with three pieces. There are two solutions.

<h3>Domains</h3>

The absolute value function on the left has a turning point where its value is zero:

1/2b -8 = 0

b -16 = 0

b = 16

The absolute value function on the right has a turning point where its value is zero:

1/4b -1 = 0

b -4 = 0

b = 4

For b > 16, both absolute value functions are identity functions. In this domain, the equation is ...

1/2b -8 = 1/4b -1

For 4 < b < 16, the function on the left negates its argument, so the equation in this domain is ...

-(1/2b -8) = 1/4b -1

For b < 4, both functions negate their arguments, so the equation in this domain is ...

-(1/2b -8) = -(1/4b -1)

For the purpose of finding the value of b, this is effectively identical to the equation for b > 16. (The value of b does not change if we multiply both sides of the equation by -1.)

<h3>Solutions</h3>

<u>Domain b < 4 ∪ 16 < b</u>

1/2b -8 = 1/4b -1

2b -32 = b -4 . . . . . . . . multiply by 4

b = 28 . . . . . . . . . . . . add 32-b to both sides

This solution is in the domain of the equation, so is one of the solutions to the original equation.

<u>Domain 4 < b < 16</u>

-(1/2b -8) = 1/4b -1 . . . . equation in this domain

-2b +32 = b -4 . . . . . . multiply by 4

36 = 3b . . . . . . . . . . . add 2b+4 to both sides

12 = b . . . . . . . . . . . . divide by 3

This solution is in the domain of the equation, so is the other solution to the original equation.

<h3>Graph</h3>

For the purposes of the graph, we have define the function g(b) to be the difference of the two absolute value functions. The solutions are found where g(x) = 0, the x-intercepts. The graph shows those to be ...

b = 12 and b = 28

<em>Additional comment</em>

Defining g(b) = |1/2b-8| -|1/4b-1|, we can rewrite it as ...

$g(b)=\begin{cases}7-\dfrac{1}{4}b&\text{for }b < 4\\-\dfrac{3}{4}b+9&\text{for }4\le b < 16\\\dfrac{1}{4}b-7&\text{for }16\le b\end{cases}$

Then the solutions are the values of b that make g(b) = 0.

4 0

2 years ago