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3 years ago

Factoring trinomials: 8)-2x2 + 8x + 10 *

Mathematics

2 answers:

lina2011 [118]3 years ago

8 0

Answer:

-2(x - 5)(x + 1)

Step-by-step explanation:

-2x² + 8x + 10

-2(x² - 4x - 5)

-2(x - 5)(x + 1)

frutty [35]3 years ago

3 0

Step-by-step explanation:

Hey, there!!

Let's do it simply, ok.

Given that,

$= - 2 {x}^{2} + 8x + 10$

Take (-2) common to make it easy.

$= -2( 2 {x}^{2} - 8x - 10)$

= -2{ x^2-(5-1)x -5}

Multily with x,

= -2 {x^2-5x+x-5}

Taking common,

= -2{x(x-5) +1(x-5)

= -2{(x+1)(x-5)}...is your required answer.

Hope it helps...

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A_____plot is used to display and summarize data using quartiles and median

NARA [144]

Answer:

BOX

Step-by-step explanation:

5 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

What is the value of x? Enter your answer in the box.

uysha [10]

From the picture you can see that TY bisects angle T.

Use the bisector property:

$\dfrac{TK}{KY} =\dfrac{TV}{VY}$ .

If YK=68 mm and VK=x mm, then VY=x-68 mm.

Then $\dfrac{129.2}{68} =\dfrac{57}{x-68}$ .

Express x-68:

$x-68=\dfrac{57\cdot 68}{129.2}=30$ mm.

Therefore, x=30+68=98 mm.

Answer: x=98 mm.

6 0

2 years ago

Find the measure of the exterior angle.

Marysya12 [62]

Answer:

The exterior angle is 77

Step-by-step explanation:

The measure of the exterior angle of a triangle is equal to the sum of the two opposite interior angles.

2x+11 = x +44

First we need to find x

Subtract x from each side

2x+11-x = x+44

x +11 = 44

Subtract 11 from each side

x+11-11 = 44-11

x =33

Now we can find the exterior angle

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The exterior angle is 77

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