What is the slope of the line shown? On a coordinate plane, a line goes through (negative 4, negative 3) and (2, 0).

A. {0,1,2,3,4}<br> B. {1,2,3,4}<br> C. {-1,0,1,2,3}<br> D. {-2,-1,0,1,2,3,4}

Harlamova29_29 [7]

Option D:

The set is {-2, -1, 0, 1, 2, 3, 4}.

Solution:

Given set:

$\{x \in Z:-2 \leq x \leq 4\}$

Z means integer values.

Integer means set of positive and negative values including zero.

$-2 \leq x \leq 4$

This means x lies between -2 and 4.

If x strictly less than -2 and greater than 4 means you don't have to include -2 and 4.
But here x less than or equal to -2 and greater than or equal to 4, so you have to include -2 and 4.

The integers are -2, -1, 0, 1, 2, 3 and 4.

So, x belongs to the set {-2, -1, 0, 1, 2, 3, 4}

Hence the given set is defined by the set {-2, -1, 0, 1, 2, 3, 4}.

Option D is the correct answer.

6 0

3 years ago

Eighty-three and seventy-four hundredths minus sixty-one and ninety-seven hundredths equals _____.

beks73 [17]

21.77 you can also type that in in the google search bar and your answer will pop up

6 0

3 years ago

Read 2 more answers

Maggie graphed the image of a 90 counterclockwise rotation about vertex A of . Coordinates B and C of are (2, 6) and (4, 3) and

Lemur [1.5K]

Answer:

A(2,2)

Step-by-step explanation:

Let the vertex A has coordinates $(x_A,y_A)$

Vectors AB and AB' are perpendicular, then

$\overrightarrow {AB}=(2-x_A,6-y_A)\\ \\\overrightarrow {AB'}=(-2-x_A,2-y_A)\\ \\\overrightarrow {AB}\perp\overrightarrow {AB'}\Rightarrow \overrightarrow {AB}\cdot \overrightarrow {AB'}=0\Rightarrow (2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0$

Vectors AC and AC' are perpendicular, then

$\overrightarrow {AC}=(4-x_A,3-y_A)\\ \\\overrightarrow {AC'}=(1-x_A,4-y_A)\\ \\\overrightarrow {AC}\perp\overrightarrow {AC'}\Rightarrow \overrightarrow {AC}\cdot \overrightarrow {AC'}=0\Rightarrow (4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0$

Now, solve the system of two equations:

$\left\{\begin{array}{l}(2-x_A)(-2-x_A)+(6-y_A)(2-y_A)=0\\ \\(4-x_A)(1-x_A)+(3-y_A)(4-y_A)=0\end{array}\right.\\ \\\left\{\begin{array}{l}-4-2x_A+2x_A+x_A^2+12-6y_A-2y_A+y^2_A=0\\ \\4-4x_A-x_A+x_A^2+12-3y_A-4y_A+y_A^2=0\end{array}\right.\\ \\\left\{\begin{array}{l}x_A^2+y_A^2-8y_A+8=0\\ \\x_A^2+y_A^2-5x_A-7y_A+16=0\end{array}\right.$

Subtract these two equations:

$5x_A-y_A-8=0\Rightarrow y_A=5x_A-8$

Substitute it into the first equation:

$x_A^2+(5x_A-8)^2-8(5x_A-8)+8=0\\ \\x_A^2+25x_A^2-80x_A+64-40x_A+64+8=0\\ \\26x_A^2-120x_A+136=0\\ \\13x_A^2-60x_A+68=0\\ \\D=(-60)^2-4\cdot 13\cdot 68=3600-3536=64\\ \\x_{A_{1,2}}=\dfrac{60\pm8}{2\cdot 13}=\dfrac{34}{13},2$