All categories

3 years ago

I do not understand what subscript numbers mean

Mathematics

1 answer:

alekssr [168]3 years ago

6 0

Answer:

subscripts identify different instances of objects that have a similar meaning or definition

Step-by-step explanation:

You apparently are interested in the meaning of the subscript N in the explicit formula ...

$A_N=\dfrac{7N}{N+5}$

Here, it is used to identify different terms of a sequence. When a value is substituted for N, it replaces all instances of N. Using your example, the first three values of $A_N$ are ...

$\text{for N=1: }A_1=\dfrac{7\cdot 1}{1+5}=\dfrac{7}{6}\\\\\text{for N=2: }A_2=\dfrac{7\cdot 2}{2+5}=\dfrac{14}{7}=2\\\\\text{for N=3: }A_3=\dfrac{7\cdot 3}{3+5}=\dfrac{21}{8}$

So, if you're asked to find the 20th term of the sequence, you put 20 where N is, then do the arithmetic.

Subscripts are also used to identify elements of a matrix or coordinates in multiple dimensions. For this purpose, there may be 2 (or more) subscripts, whose order has meaning. The subscripts 24, for example, in $A_{24}$ refer to row 2, column 4 of a matrix named A. That is, the subscripts are in row, column order.

In short, subscripts are used to identify different instances of similar objects.

You might be interested in

The diameter of a circle is 18 meters. What is the circle's area?

Anarel [89]

Answer:

254.45 meters squared

Step-by-step explanation:

Well the radius is 9 and that squared is 81 then that times pi is 254.45.

3 0

4 years ago

Help please!!!!!!!!!!!!!!!!!!

gtnhenbr [62]

What do you need help on?

3 0

3 years ago

Quick algebra 1 question for 50 points!

erik [133]

Answer:

10 points

Step-by-step explanation:

From the passage, we know:

1. Jay earned 15 points after spending $60

2. We want to know the number of points he can earn if he spends $40

We can set this up into proportions.

60:15 = 40:x

The x is the number of points he will get after spending $40.

15(40) = 60x

600 = 60x

x = 10

Jay will earn 10 points if he uses $40.

4 0

2 years ago

Help with the question plz people

erastovalidia [21]

You are cheating. I see this is a test.

4 0

3 years ago

Solve these linear equations by Cramer's Rules Xj=det Bj / det A:

timurjin [86]

Answer:

(a) $x_1=-2,x_2=1$

(b) $x_1=\frac{3}{4} ,x_2=-\frac{1}{2} ,x_3=\frac{1}{4}$

Step-by-step explanation:

(a) For using Cramer's rule you need to find matrix $A$ and the matrix $B_j$ for each variable. The matrix $A$ is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get $A$ more easily.

$2x_1+5x_2=1\\x_1+4x_2=2$

$\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right]$

To get $B_1$ , replace in the matrix A the 1st column with the results of the equations:

$B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]$

To get $B_2$ , replace in the matrix A the 2nd column with the results of the equations:

$B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]$

Apply the rule to solve $x_1$ :

$x_1=\frac{det\left(\begin{array}{cc}1&5\\2&4\end{array}\right)}{det\left(\begin{array}{cc}2&5\\1&4\end{array}\right)} =\frac{(1)(4)-(2)(5)}{(2)(4)-(1)(5)} =\frac{4-10}{8-5}=\frac{-6}{3}=-2\\x_1=-2$

In the case of B2, the determinant is going to be zero. Instead of using the rule, substitute the values of the variable $x_1$ in one of the equations and solve for $x_2$ :

$2x_1+5x_2=1\\2(-2)+5x_2=1\\-4+5x_2=1\\5x_2=1+4\\ 5x_2=5\\x_2=1$

(b) In this system, follow the same steps,ust remember $B_3$ is formed by replacing the 3rd column of A with the results of the equations:

$2x_1+x_2 =1\\x_1+2x_2+x_3=0\\x_2+2x_3=0$

$\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right]$

$B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right]$

$B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right]$

$B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]$

$x_1=\frac{det\left(\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{1(2)(2)+(0)(1)(0)+(0)(1)(1)-(1)(1)(1)-(0)(1)(2)-(0)(2)(0)}{(2)(2)(2)+(1)(1)(0)+(0)(1)(1)-(2)(1)(1)-(1)(1)(2)-(0)(2)(0)}\\ x_1=\frac{4+0+0-1-0-0}{8+0+0-2-2-0} =\frac{3}{4} \\x_1=\frac{3}{4}$

$x_2=\frac{det\left(\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{(2)(0)(2)+(1)(0)(0)+(0)(1)(1)-(2)(0)(1)-(1)(1)(2)-(0)(0)(0)}{4} \\x_2=\frac{0+0+0-0-2-0}{4}=\frac{-2}{4}=-\frac{1}{2}\\x_2=-\frac{1}{2}$

$x_3=\frac{det\left(\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)}=\frac{(2)(2)(0)+(1)(1)(1)+(0)(1)(0)-(2)(1)(0)-(1)(1)(0)-(0)(2)(1)}{4} \\x_3=\frac{0+1+0-0-0-0}{4}=\frac{1}{4}\\x_3=\frac{1}{4}$

6 0

4 years ago