Using it's concept, the domain of the function shown in the graph is:
B. 
.
<h3>What is the domain of a function?</h3>
The domain of a function is the set that contains all possible input values for the function. In a graph, the domain is the set that contains all values of x.
From this graph, the function is defined for values of x between -6 and 6, hence the domain is given by:
B. .
More can be learned about the domain of a function at brainly.com/question/10891721
#SPJ1
To find the percentage of 2/5 you'll divide them and then multiply the answer by 100.
2/5 = 0.4 = 40%
So 40% and 2/5 are equal.
Hope this helps :)
Refer to the attachment for the visualization of the staircase that is 4 units tall and contains 10 unit squares.
This problem can easily be solved by using the formula for
arithmetic series. The formula is given by:
![S_{n}= \frac{n}{2}[2 a_{1}+(n-1)d]](https://tex.z-dn.net/?f=%20S_%7Bn%7D%3D%20%5Cfrac%7Bn%7D%7B2%7D%5B2%20a_%7B1%7D%2B%28n-1%29d%5D%20%20)
where S is the sum, n is the number of terms,
is the first term, and d is the common difference.
In the case of the problem, n is 4, the first term is 1, and the common difference is also 1. If you substitute these conditions to the formula, you'll get 10 square units.
Now for the extended staircase, we will have 12 number of terms since the staircase is now 12 units tall. The common difference and the first term will still be the same as the previous condition. We can solve for the unit squares it contains by substituting these values to the formula for arithmetic series:
![S_{12}= \frac{12}{2}[2(1)+(12-1)(1)]](https://tex.z-dn.net/?f=%20S_%7B12%7D%3D%20%5Cfrac%7B12%7D%7B2%7D%5B2%281%29%2B%2812-1%29%281%29%5D%20)
![S_{12}= 6[2+11]=6(13)=78](https://tex.z-dn.net/?f=%20S_%7B12%7D%3D%206%5B2%2B11%5D%3D6%2813%29%3D78%20)
ANSWER: The extended staircase would contain 78 unit squares.
Answer:
1 and 1
Step-by-step explanation:
Answer:
These techniques for elimination are preferred for 3rd order systems and higher. They use "Row-Reduction" techniques/pivoting and many subtle math tricks to reduce a matrix to either a solvable form or perhaps provide an inverse of a matrix (A-1)of linear equation AX=b. Solving systems of linear equations (n>2) by elimination is a topic unto itself and is the preferred method. As the system of equations increases, the "condition" of a matrix becomes extremely important. Some of this may sound completely alien to you. Don't worry about these topics until Linear Algebra when systems of linear equations (Rank 'n') become larger than 2.
Step-by-step explanation:
Just to add a bit more information, "Elimination" Can have a variety of other interpretations. Elimination techniques typically refer to 'row reduction' to achieve 'row echelon form.' Do not worry if you have not heard of these terms. They are used in Linear Algebra when referring to "Elimination techniques"
Gaussian Elimination
Gauss-Jordan Elimination
LU-Decomposition
QR-Decomposition
These techniques for elimination are preferred for 3rd order systems and higher. They use "Row-Reduction" techniques/pivoting and many subtle math tricks to reduce a matrix to either a solvable form or perhaps provide an inverse of a matrix (A-1)of linear equation AX=b. Solving systems of linear equations (n>2) by elimination is a topic unto itself and is the preferred method. As the system of equations increases, the "condition" of a matrix becomes extremely important. Some of this may sound completely alien to you. Don't worry about these topics until Linear Algebra when systems of linear equations (Rank 'n') become larger than 2.
Substitution is the preferred method for 2 equations in 2 unknowns. The constants are unimportant other than having a non-zero determinant. It is always easy to find multiplicative factors using LCMs of one variable or the other to allow substitution into the other equation:
Example:
4X + 5Y = 9
5X - 4Y = 1
Notice that 20 is a LCM of either the X or Y variable. So multiply the first by 4 and the second by 5 and then adding the two (Y's will drop out allowing for substitution)
4(4X + 5Y = 9)
5(5X - 4Y = 1)
Multiplying to produce the LCM factors:
16X + 20Y = 36
25X - 20Y = 5
Adding the equations
41X = 41
X = 1
Substitution into either equation yields
Y = 1
Elimination techniques are preferred for Rank-n>3
