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3 years ago

Jane thought of a number then added -5 and subtracted -8 to get 13. What number did Jane start with

Mathematics

1 answer:

ad-work [718]3 years ago

3 0

Rephrased: Jane thought of a number, subtracted 5, and added 8 to get 13. -5+8=3. 13-3=10. Jane’s number is 10.

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Justin is covering the outside of an open shoe box with colorful paper for a class project. The shoebox is 30 cm long, 20 cm wid

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If lines L and M are parallel, find x, y, and z. Explain your reasoning.

Katyanochek1 [597]

Answer:

The answer to your question is:

x = 43°, y = 137° and z= 43°

Step-by-step explanation:

-If lines L and M are parallel, that means that 43° and z are corresponding angles and they measure the same.

z = 43°

- y and 43° are supplementary angles , so

y + 43 = 180

y = 180 - 43

y = 137°

- x and z are vertical angles, so they measure the same

x = z = 43°

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3 years ago

C=1/4 (d+3) make d the subject

trasher [3.6K]

Answer:

4c-3=d

Step-by-step explanation:

4c=d+3

4c-3=d

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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.

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

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

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3 years ago

Matthew is 5 years older than twice the age of his sister. If Matthew is 13, how old is his sister? 3 4 9 31

Zepler [3.9K]

Answer:

Step-by-step explanation:

first take away the 5 years (13-5=8)

then divide it by 2 (8÷2=4

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3 years ago