(9x^4-3x^3+4x^2+5x+7) + (11x^4-4x^2-11x-9)

Can you Help me please

Evgen [1.6K]

Answer:

-1

Step-by-step explanation:

mn = -3 because 3 × -1 = -3

p^2 = 4 because 2^2 = 4

Then 4 + -3 = -1

6 0

3 years ago

Read 2 more answers

I need help on this question

Alona [7]

Answer:

1. equal

2. add to 180

3. equal

4. add to 180

Step-by-step explanation:

as 1 and 3 are linear pairs

2 and 4 are vertically opposite angles

3 0

3 years ago

Flor tiene una papelería y vende los lápices en $12 y los bolígrafos en $20. ¿Qué expresión algebraica permite saber cuánto debe

stiks02 [169]

Answer:

c=120x+200y

Step-by-step explanation:

La expresión algebraica debería indicar que el valor total que debe cobrar Flor es igual a 10 que representa el número de bolsas por el resultado de la suma del precio de cada item por la cantidad, lo que se puede expresar de la siguiente forma:

c=10(12x+20y)

c=120x+200y

De acuerdo a esto, la expresión algebraica que permite saber cuánto debe cobrar Flor por 10 bolsas que contengan x cantidad de lápices y "y" cantidad de bolígrafos es c=120x+200y.

7 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

An equation in the system is y = x + 1. What is the other equation in the system if the solution is only at (0, 1)?

Elza [17]

Answer:

D. x + 2y = 2

Step-by-step explanation:

0 + 2(1) = 2

4 0

4 years ago