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3 years ago

First (a) simplify the expression and then (b) evaluate the expression for x = -8.

Mathematics

2 answers:

PolarNik [594]3 years ago

4 0

Answer: 4

Step-by-step explanation:

a) Simplied Expresion

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

2x+5(x-4-2x)

2x+5(-x-4)

2x-5x-20

-3x-20

b) x=-8

-3(-8)-20= 4

Elza [17]3 years ago

3 0

A)
2x + 5 [x -2(2+x)]
2x + 5 (x -4 -2x)
2x +5 (-x -4)
2x -5x -20
-3x-20 => answer

B)
-3x -20
-3(-8) -20
24 -20
4 => answer

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Assume that T is a linear transformation. Find the standard matrix of T. T: R^3 right arrow R^2 , T(e 1) =(1,2), and T(e2 ) =( -

irina [24]

Answer:

$A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]$

Step-by-step explanation:

Given

$T:R^3->R^2$

$T(e_1) = (1,2)$

$T(e_2) = (-4,6)$

$T(e_3) = (2,-6)$

Required

Find the standard matrix

The standard matrix (A) is given by

$Ax = T(x)$

Where

$T(x) = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]$

$Ax = T(x)$ becomes

$Ax = [T(e_1)\ T(e_2)\ T(e_3)]\left[\begin{array}{c}x_1&x_2&x_3\\-&&x_n\end{array}\right]$

The x on both sides cancel out; and, we're left with:

$A = [T(e_1)\ T(e_2)\ T(e_3)]$

Recall that:

$T(e_1) = (1,2)$

$T(e_2) = (-4,6)$

$T(e_3) = (2,-6)$

In matrix:

$(a,b)$ is represented as: $\left[\begin{array}{c}a\\b\end{array}\right]$

So:

$T(e_1) = (1,2) = \left[\begin{array}{c}1\\2\end{array}\right]$

$T(e_2) = (-4,6)=\left[\begin{array}{c}-4\\6\end{array}\right]$

$T(e_3) = (2,-6)=\left[\begin{array}{c}2\\-6\end{array}\right]$

Substitute the above expressions in $A = [T(e_1)\ T(e_2)\ T(e_3)]$

$A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]$

Hence, the standard of the matrix A is:

$A = \left[\begin{array}{ccc}1&-4&2\\2&6&-6\end{array}\right]$

6 0

3 years ago

The midpoint of A (-4, 2) and B(8, 5) is

Norma-Jean [14]

$\bf \textit{middle point of 2 points }\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ -4}}\quad ,&{{ 2}})\quad 
% (c,d)
&({{ 8}}\quad ,&{{ 5}})
\end{array}\qquad
% coordinates of midpoint 
\left(\cfrac{{{ x_2}} + {{ x_1}}}{2}\quad ,\quad \cfrac{{{ y_2}} + {{ y_1}}}{2} \right)
\\\\\\
\left(\cfrac{{{ 8}} -4}{2}\quad ,\quad \cfrac{{{ 5}} + {{ 2}}}{2} \right)$

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3 years ago

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lozanna [386]

Answer:

A. the times at which the golf ball is on the ground

Step-by-step explanation:

The expression of the function is

h(x)= -4x^2+36x

The roots can be seen in the image below

You have a formula which represents the height over time.

The roots of the equation indicate that the height is equal to zero

The correct option is

A. the times at which the golf ball is on the ground

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3 years ago

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Vlad1618 [11]

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Substitute in y for either original equation.
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dlinn [17]

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