Find the gcf of 11 and 13

Mathematics

2 answers:

photoshop1234 [79]2 years ago

8 0

The answer to this question is the gcf is 1

jarptica [38.1K]2 years ago

3 0

GCF = greatest common factor in both terms.
So to find out the gcf let's write some factors of each

11 => 1 * 11 = 11
=> 11 * 1 = 11
So basically the factors are 1 and 11

13 => 1 * 13 = 13
=> 13 * 1 = 13
factors are 1 and 13

So now we can see that the number which is divisible by both are 1
So, GCF of 11 and 13 is 1.

Hope this helped.. :)

You might be interested in

What is a budget? only a savings plan savings and spending plan only a spending plan

son4ous [18]

Answer:

Step-by-step explanation:

A plan to only spend x amount of money while shopping

7 0

3 years ago

Read 2 more answers

Please simplify, 50 points.

olga55 [171]

Answer:

The trigonometric equation (sin Θ − cos Θ)^2 − (sin Θ + cos Θ)^3 can be simplified by:Using x for Θ: (sinx - cosx)^2 - (sinx + cosx)^2 = (sin^2 x - 2sinxcosx + cos^2 x) - (sin^2 x + 2sinxcosx + cos^2 x) = - 2 sinx cosx - 2 sinx cosx = - 4 sinx cosx = - 2sin(2x)

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=81-%28y-1%29%28y%5E2%2B27%29%20%5Cgeq%200%20%5C%5C%5C%5C%2081-%28y%5E3%2B27y-y%5E2-27%29%20%5C

Zarrin [17]

First thing you'll need to know is that the value for this equation is actually an approximation 'and' it is imaginary, so, one method is via brute force method.

You let f(y) equals to that equation, then, find the values for f(y) using values from y=-5 to 5, you just substitute the values in you'll get -393,-296,-225,... till when y=3 is f(y)=-9; y=4 is f(y)=48, so there is a change in signs when 'y' went from y=3 to y=4, the answer is between 3 and 4, you can work out a little bit deeper using 3.1, 3.2... You get the point. The value is close to 3.1818...

The other method is using Newton's method, it is similar to this but with a twist because it involves differentiation, so $y_{n+1}=y_n-\frac{f(y)}{f'(y)}$ where 'n' is the number you approximate, like n=0,1,2... etc. f(y) would the equation, and f'(y) is the derivative of f(y), now what you'll need to do is substitute the 'n' values into 'y' to find the approximation.