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3 years ago

PLEASE HELP Factor the common factor out of each expression. 1) 16x -6

Mathematics

1 answer:

zavuch27 [327]3 years ago

8 0

Answer:

8x-3

Step-by-step explanation:

take a 2 out of each of them

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B. The 100th term of 60, 110, 160, ... is (Simplify your answer.) How to do this?

Advocard [28]

The 1st term is 60.

Add 50 to this to get the 2nd term, 60 + 50 = 110.

Add 50 to that to get the 3rd term, 110 + 50 = 160.

Add 50 to that to get the 4th term, 160 + 50 = 210.

And so on...

Notice that in the 2nd term, we added 1 copy of 50 to the 1st term.

In the 3rd, we ultimately added 2 copies of 50 to the 1st term.

In the 4th, we added 3 50s.

And so on... If the pattern continues, then the <em>n</em>-th term can be obtained by adding (<em>n</em> - 1) copies of 50 to the first term.

So, the 100th term is

60 + (100 - 1) * 50 = 5010

5 0

3 years ago

ASAPPP!!!!!

Sergio039 [100]

Answer:

185

Step-by-step explanation:

45 +70(2) = 185

5 0

4 years ago

Please help, love you

ExtremeBDS [4]

Answer:

162

Step-by-step explanation:

Note that you can split the given shape into a triangle and a rectangle.

Area of rectangle = base x height

base = 12

height = 9

Plug in the corresponding numbers into the corresponding variables:

Area of rectangle = 12 x 9

= 108

Area of triangle = (base x height)/2

base = 12

height = 9

Area of triangle = (12 x 9)/2

= (108)/2

= 54

Next, add the two areas together to get the full area:

108 + 54 = 162

162 is your answer.

5 0

3 years ago

A sign in a bakery gives these options: • 12 cupcakes for $29 • 24 cupcakes for $56 • 50 cupcakes for $129 Find each unit price

crimeas [40]

Answer:

a) $2.44 per cupcake

b) $2.33 per cupcake

c) $2.58 per cupcake

Step-by-step explanation:

We are given the following in the question:

We have to calculate the unit price of the cupcakes.

Unit price =

$=\dfrac{\text{Total Cost}}{\text{Number of cupcakes}}$

12 cupcakes for $29

Unit price =

$=\dfrac{29}{12}=2.43\text{ Dollar per cupcake}$

24 cupcakes for $56

Unit price =

$=\dfrac{56}{24}=2.33\text{ Dollar per cupcake}$

50 cupcakes for $129

Unit price =

$=\dfrac{129}{50}=2.58\text{ Dollar per cupcake}$

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