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3 years ago

4. Michael observed he felt the pain of losing a $20 bill more than he felt the joy of finding it on the sidewalk the

Mathematics

1 answer:

trasher [3.6K]3 years ago

4 0

Answer:

Endowment effect

Step-by-step explanation:

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State the additive property of zero using the variable b.

Juli2301 [7.4K]

Answer with Step-by-step explanation:

We are given a variable b

We have to state the additive property of zero using the variable b.

Additive property of zero: It states that when any number b is added to zero then we get sum is equal to number itself.

Mathematical representation:

$b+0=0+b=b$

Suppose, a number b=9

Then, 9+0=9

0+9=9

This property is called additive property of zero because when 9 is added to 0 then we get sum equals to 9.

7 0

3 years ago

-4.8=1.2s+2s+4 Please help me

nikdorinn [45]

Simplify both sides of the equation
-4.8=1.2s+2s+4
Combine like terms
-4.8=(1.2s+2s)+(4)
-4.8=3.2s+4
Flip the equation
3.2s+4=-4.8
Subtract both sides by 4
3.2s+4-4=-4.8-4
3.2s=-8.8
Divide both sides by 3.2
3.2s/3.2= -8.8/3.2
S=-2.75 will be the answer hope it helps

6 0

3 years ago

Which of the following is an irrational number? A 2.145 B -3 C 0 D π .

RoseWind [281]

Answer:

D. is the correct answer

6 0

3 years ago

Read 2 more answers

What is the answer to a/-2 = 8

WARRIOR [948]

Answer:

-16

Step-by-step explanation:

a/-2=8

you multiply -2 on both side a/-2(-2)=8(-2)

-2 cancels out one the one side

your final answer is a= -16

3 0

3 years ago

Determine if the columns of the matrix form a linearly independent set. Justify your answer. [Start 3 By 4 Matrix 1st Row 1st Co

Volgvan

Answer:

Linearly Dependent for not all scalars are null.

Step-by-step explanation:

Hi there!

1)When we have vectors like $v_{1},v_{2},v_{3}, ...$ we call them linearly dependent if we have scalars $a_{1},a_{2},a_{3},...$ as scalar coefficients of those vectors, and not all are null and their sum is equal to zero.

$a_{1}\vec{v_{1}}+a_{2}\vec{v_{2}}+a_{3}\vec{v_{3}}+...a_{m}\vec{v_{m}}=0$

When all scalar coefficients are equal to zero, we can call them linearly independent

2) Now let's examine the Matrix given:

$\begin{bmatrix}1 &-2 &2 &3 \\ -2 & 4 & -4 &3 \\ 0&1 &-1 & 4\end{bmatrix}$

So each column of this Matrix is a vector. So we can write them as:

$\vec{v_{1}}=\left \langle 1,-2,1 \right \rangle,\vec{v_{2}}=\left \langle -2,4,-1 \right \rangle,\vec{v_{3}}=\left \langle 2,-4,4 \right \rangle\vec{v_{4}}=\left \langle 3,3,4 \right \rangle$ Or

Now let's rewrite it as a system of equations:

$a_{1}\begin{bmatrix}1\\ -2\\ 0\end{bmatrix}+a_{2}\begin{bmatrix}-2\\ 4\\ 1\end{bmatrix}+a_{3}\begin{bmatrix}2\\ -4\\ -1\end{bmatrix}+a_{4}\begin{bmatrix}3\\ 3\\ 4\end{bmatrix}=\begin{bmatrix}0\\ 0\\ 0\end{bmatrix}$

2.1) Since we want to try whether they are linearly independent, or dependent we'll rewrite as a Linear system so that we can find their scalar coefficients, whether all or not all are null.

Using the Gaussian Elimination Method, augmenting the matrix, then proceeding the calculations, we can see that not all scalars are equal to zero. Then it is Linearly Dependent.

$\left ( \left.\begin{matrix}1 &-2 &2 &3 \\ -2 &4 &-4 &3 \\ 0 & 1 &-1 &4 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )R_{1}\times2 +R_{2}\rightarrow R_{2}\left ( \left.\begin{matrix}1 &-2 &2 &3 \\ 0 &0 &9 &0\\ 0 & 1 &-1 &4 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )\ R_{2}\Leftrightarrow R_{3}\left ( \left.\begin{matrix}1 &-2 &2 &3 \\ 0 &1 &-1 &4 \\ 0 &0 &9 &0 \\ \end{matrix}\right|\begin{matrix}0\\ 0\\ 0\end{matrix} \right )$ $\left\{\begin{matrix}1a_{1} &-2a_{2} &+2a_{3} &+3a_{4} &=0 \\ &1a_{2} &-1a_{3} &+4a_{4} &=0 \\ & & &9a_{4} &=0 \end{matrix}\right.\Rightarrow a_{1}=0, a_{2}=a_{3},a_{4}=0$

$S=\begin{bmatrix}0\\ a_{3}\\ a_{3}\\ 0\end{bmatrix}$

3 0

3 years ago