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2 years ago

Solve the inequality 7+7(x+4)>21

Mathematics

1 answer:

Kryger [21]2 years ago

8 0

Answer:

x>-2

Step-by-step explanation:

Hi there!

We are given the inequality 7+7(x+4)>21, and we want to solve it

Solving an inequality is a lot like solving an equation; we want to isolate x on one side by itself.

So to start, we'll preform the distributive property on 7(x+4) (multiply x by 7, then multiply 4 by 7 and add those two together).

7+7x+28>21

Now combine like terms.

7x+35>21

Subtract 35 from both sides

7x>-14

Divide both sides by 7

x>-2

Hope this helps!

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5x+6y=7 need to find the slope intercept and standard form

Oksi-84 [34.3K]

The problem is already in standard form but to get it into slope intercept form you rewrite the equation so it looks like
6y=-5x+7 divide all by 6
y=-5/6x+7/6
slope being -5/6 and b being 7/6

7 0

2 years ago

How do you perform this: 2x+3y=25 3x+4y=22

meriva

Answer:

x = -34 , y = 31

Step-by-step explanation suing Gaussian elimination:

Solve the following system:

{2 x + 3 y = 25

3 x + 4 y = 22

Express the system in matrix form:

(2 | 3

3 | 4)(x

y) = (25

22)

Write the system in augmented matrix form and use Gaussian elimination:

(2 | 3 | 25

3 | 4 | 22)

Swap row 1 with row 2:

(3 | 4 | 22

2 | 3 | 25)

Subtract 2/3 × (row 1) from row 2:

(3 | 4 | 22

0 | 1/3 | 31/3)

Multiply row 2 by 3:

(3 | 4 | 22

0 | 1 | 31)

Subtract 4 × (row 2) from row 1:

(3 | 0 | -102

0 | 1 | 31)

Divide row 1 by 3:

(1 | 0 | -34

0 | 1 | 31)

Collect results:

Answer: {x = -34 , y = 31

6 0

2 years ago

How do you do this haaaa

Ratling [72]

Answer:

7000

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

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.

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