All categories

3 years ago

Which two of the following are applications of the Logarithmic model? *

Mathematics

1 answer:

sukhopar [10]3 years ago

7 0

Answer:

Richter Scale and Acidity/Alkalinity

Step-by-step explanation:

College GPA is modelled by weighted means, Compound Interest uses a potential model, amortization schedule is based on succesions, whereas Richter Scale and Acidity/Alkalinity use logartihmic model.

You might be interested in

The number expressed using exponents

Anna007 [38]

This expression can be written in a shorter way using something called exponents. An expression that represents repeated multiplication of the same factor is called a power.The number 5 is called the base, and the number 2 is called the exponent. For example $x^{2}$

8 0

4 years ago

Unite price batteries size 4pk and total price 2.97 how come unite price

omeli [17]

To find the unit price, divide the unit by the the total amount:

2.97/4
0.7425

Round this to the nearest cent:

$0.74

Hope this helps!

5 0

3 years ago

Read 2 more answers

Can someone help me? I don’t really understand how to do this m

Artemon [7]

Answer:

probably about 59 degrees

Step-by-step explanation:

4 0

4 years ago

Read 2 more answers

2,161 as a percentage?

kramer

Answer:

Step-by-step explanation:

24.52

4 0

4 years ago

find a homogeneous linear system of two equations in three unknowns whose solution space consists of those vectors in R3 that ar

Leona [35]

$\mathbf a=(1,1,1)^\top=1\,\mathbf i+1\,\mathbf j+1\,\mathbf k$
$\mathbf b=(-2,3,0)^\top=-2\,\mathbf i+3\,\mathbf j$
$\mathbf a\times\mathbf b=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\1&1&1\\-2&3&0\end{vmatrix}=-3\,\mathbf i-2\,\mathbf j+5\,\mathbf k=(-3,-2,5)^\top$

Basically, you're looking for a matrix $\mathbf A$ such that

$\mathbf A(\mathbf a\times\mathbf b)=\mathbf 0$

i.e. a matrix $\mathbf A$ whose nullspace with basis vector $\mathbf a\times\mathbf b$ .

By the rank-nullity theorem, the rank of $\mathbf A$ and the dimension of its nullspace must add up to the number of columns, so

$\mathrm{rank}\mathbf A+\underbrace{\mathrm{null}\mathbf A}_1=3\implies\mathrm{rank}\mathbf A=2$

One easy choice for a row would be $\begin{bmatrix}1&1&1\end{bmatrix}$ , since

$(1,1,1)(-3,-2,5)^\top=0$

Now you only need to find another combination such that the second row of $\mathbf A$ is independent of the first. An easy choice for this is to let the first element be 0, and the next be 1. Then the last element must be $\dfrac25$ , as

$\left(0,1,\dfrac25\right)(-3,-2,5)^\top=0$

So,

$\underbrace{\begin{bmatrix}1&1&1\\0&1&\frac25\end{bmatrix}}_{\mathbf A}\begin{bmatrix}-3\\-2\\5\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$

is one possible solution.

3 0

3 years ago