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3 years ago

Determine area of the peice

Mathematics

2 answers:

Rasek [7]3 years ago

8 0

79.56 is the answer.

katen-ka-za [31]3 years ago

3 0

The answer is 79.56.

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What are the solutions to 3n-1 > 8 or 4n+ 3

Mandarinka [93]

Answer:

n>3

Step-by-step explanation:

Step 1: Add 1 to both sides.

3n−1+1>8+1

3n>9

Step 2: Divide both sides by 3.

3n/3>9/3

n>3

3 0

3 years ago

Read 2 more answers

HEEELLLPP PLEASEE I GIVE BRAINLIST!!!

Mila [183]

Take any of the valid pairs of points, use the formula
rate of change = (y2-y1)/(x2-x1)

Say we use P1=(0,21.4), P2=(3,19)
x1=0, y1=21.4
x2=3, y2=19

Rate of change = (21.4-19)/(0-3)=2.4/(-3) = -0.8 sq. ft/in of liquid

4 0

3 years ago

The cell phone company decided to collect similar data but this time they collected information of texting speed according to ag

Ludmilka [50]

Step-by-step explanation:

As we know, Scatter plots resembles line graphs in that they are plotted on the x and y axis and the purpose of a scatter plot is to shows if the variables are related to each other. If yes, they can draw a line of best fit which can expose or go through most of the points in the plot.

Given the information in the question, we know:

"they collected information of texting speed according to age"

=> We have 2 variables here which are age and (texting speed)

Step 1:

Label the x-axis the input variable (age)
Label the y-axis the output variable (texting speed)

Step 2: Plot the points according to age and texting speed

Hence, from that plot, it could have two cases in the relation of age and texting speed which are "the more time, more text is written", or "the less time, more text is written".

7 0

3 years ago

Read 2 more answers

Cara and her brother take a numbered spinner from a board game and use it to determine

Oksi-84 [34.3K]

They are dependent events because if cara spins it and it lands on her she gets to play with the puppy, But if she misses it and her brother spins and it lands on him he gets to play with the puppy.

Short answer: the event is dependent because the outcome of the first spin effects the brother outcome

4 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