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andreev551 [17]

3 years ago

12

Define an operation ‘&’ on the set of positive integers such that: 2 & 1 = 5 3 & 2 = 13 5 & 3 = 34

1 answer:

Scilla [17]3 years ago

7 0

Answer:What are the equivalence classes of the equivalence relations in Exercise 3? A binary relation defined on a set S is said to be equivalence relation if it is reflexive, symmetric and transitive. An equivalence relation defined on a set S, partition the set into disjoint equivalence classes

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Evaluate each value expression for the given value of x: x-5 , x - =22<br> pls fast

masya89 [10]

Answer:

-18

Step-by-step explanation:

-18

brainlest please

6 0

3 years ago

The point-slope form of the equation of a line that passes through points (8, 4) and (0, 2) is y – 4 = y minus 4 equals StartFra

Stels [109]

Answer:

Third option: $y= \frac{1}{4}x+2$

Step-by-step explanation:

<h3><em> The correct form of the exercise is: "The point-slope form of the equation of a line that passes through points (8, 4) and (0, 2) is $y -4 = \frac{1}{4}(x -8)$ . What is the slope-intercept form of the equation for this line?"</em></h3><h3><em /></h3>

<em> </em>The equation of the line in Slope-Intercept form is:

$y=mx+b$

Where "m" is the slope and "b" is the y-intercept.

Given the equation of the line in Point-Slope form:

$y -4 = \frac{1}{4}(x -8)$

You need to solve for "y" in order to write the given equation of the line in Slope Intercept form.

Then, this is:

$y -4 = \frac{1}{4}(x -8)\\\\y-4= \frac{1}{4}x-\frac{8}{4}\\\\y-4= \frac{1}{4}x-2\\\\y= \frac{1}{4}x-2+4\\\\y= \frac{1}{4}x+2$

You can identify that the slope "m" is:

$m=\frac{1}{4}$

And the y-interecept "b" is:

$b=2$

7 0

3 years ago

Read 2 more answers

Find the slope of the line. <br><br> A. 1/3<br> B. - 1/3<br> C. 3<br> D. -3

BARSIC [14]

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have the points (0, -2) and (6, 0).

Substitute:

$m=\dfrac{0-(-2)}{6-0}=\dfrac{2}{6}=\dfrac{1}{3}$

<h3>Answer: A. 1/3</h3>

3 0

3 years ago

Suppose that a function f(x) is defined for all real values of x except at xequals=c. can anything be said about the existence

Margaret [11]

we are given that

f(x) is defined for all values of x except at x=c

Limit may or may not exist

case-1:

If there is hole at x=c , then limit exist

case-2:

If there is vertical asymptote at x=c , then limit does not exist

Examples:

case-1:

$\lim_{x \to c} \frac{x^2-cx}{(x-c)}$

We can simplify it

$\lim_{x \to c} \frac{x(x-c)}{(x-c)}$

$=\lim_{x \to c} x$

$=c$

so, we can see that limit exist and it's value defined

case-2:

$\lim_{x \to c} \frac{1}{(x-c)}$

Left limit is

$\lim_{x \to c-} \frac{1}{(x-c)}$

$=-\infty$

Right Limit is

$\lim_{x \to c+} \frac{1}{(x-c)}$

$=+\infty$

so, we can see that left limit is not equal to right limit

so, limit does not exist

4 0

3 years ago

A particular plant root grows 3.5 inches per month. How many centimeters is the plant root growing per month? (1 inch = 2.54 cen

Dahasolnce [82]

Answer:

8.89 centimeters

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers