8 -4 -3
3 -9 -5-8 8 8
This is a matrix of 3 rows and 3 columns (it's called square matrix because the number of rows = the number of columns)
To localise an element of a matrix we use indices R and C, the first index being ALWAYS the row and the second, ALWAYS the column.
Hence:
A₂₃ = the element in row 2 and column 3, that is - 5
COLUMN
1st 2nd 3rd
--------------------------------
1st | 8 -4 -3
ROW 2nd | 3 -9 -5 3rd |-8 8 8
Answer:
m<2=133°
Step-by-step explanation:
180-47=133
Hope this helps! :)
Option B:
is the equivalent expression to the given expression.
Solution:
Given expression:
![$\sqrt[4]{9} ^{\frac{1}{2}x}](https://tex.z-dn.net/?f=%24%5Csqrt%5B4%5D%7B9%7D%20%5E%7B%5Cfrac%7B1%7D%7B2%7Dx%7D)
To find the equivalent expression to the given expression.
![$\sqrt[4]{9} ^{\frac{1}{2}x}](https://tex.z-dn.net/?f=%24%5Csqrt%5B4%5D%7B9%7D%20%5E%7B%5Cfrac%7B1%7D%7B2%7Dx%7D)
Using radical rule: ![$\sqrt[n]{a}=a^{\frac{1}{n}}](https://tex.z-dn.net/?f=%24%5Csqrt%5Bn%5D%7Ba%7D%3Da%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D)
So that
.
![$\sqrt[4]{9} ^{\frac{1}{2}x}=\left(9^{\frac{1}{4}}\right)^{\frac{1}{2} x}](https://tex.z-dn.net/?f=%24%5Csqrt%5B4%5D%7B9%7D%20%5E%7B%5Cfrac%7B1%7D%7B2%7Dx%7D%3D%5Cleft%289%5E%7B%5Cfrac%7B1%7D%7B4%7D%7D%5Cright%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%20x%7D)
Using exponent rule: 


![$\sqrt[4]{9} ^{\frac{1}{2}x}=9^{\frac{1}{8}x}](https://tex.z-dn.net/?f=%24%5Csqrt%5B4%5D%7B9%7D%20%5E%7B%5Cfrac%7B1%7D%7B2%7Dx%7D%3D9%5E%7B%5Cfrac%7B1%7D%7B8%7Dx%7D)
Hence
is the equivalent expression to the given expression.
Option B is the correct answer.
Answer: Proportional is two varying quantities are said to be in a relation of proportionality, if they are multiplicatively connected to a constant, that is, when either their ratio or their product yields a constant. The value of this constant is called the coefficient of proportionality or proportionality constant.
First, recall that Gaussian quadrature is based around integrating a function over the interval [-1,1], so transform the function argument accordingly to change the integral over [1,5] to an equivalent one over [-1,1].



So,

Let

. With

, we're looking for coefficients

and nodes

, with

, such that

You can either try solving for each with the help of a calculator, or look up the values of the weights and nodes (they're extensively tabulated, and I'll include a link to one such reference).
Using the quadrature, we then have
