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2 years ago

What is the factored form of x^2-x-2

Mathematics

1 answer:

lisov135 [29]2 years ago

5 0

Answer:

(x-2),(x+1)

Step-by-step explanation:

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Which equation describes the graph?

Leya [2.2K]

y= |2x|-1

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3 years ago

Answer Answer Answer Answer Answer Answer Answer Answer Answer Answer Answer Answer Answer Answer Answer Answer Answer Answer An

hram777 [196]

Answer:

They are both negative

Step-by-step explanation:

the given coordinate is in quadrant IV and when reflected over the y axis, quadrant IV becomes quadrant III which is (-,-)

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2 years ago

Read 2 more answers

4 Ayla s is paid $15 20 per hour for a 34 hour week. April has the same weekly pay but works 38

VashaNatasha [74]

Find Atlas weekly pay by multiplying rate by hours:

15.20 x 34 = 516.80 a week.

Now to find Apri’s rate divide the weekly pay by hours:

516.80 / 38 =$13.60 per hour

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3 years ago

<img src="https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Bx%2By%3D1%7D%20%5Catop%20%7Bx-2y%3D4%7D%7D%20%5Cright.%20%5C%5C%5Clef

brilliants [131]

Answer:

(a) x=2, y=-1

(b) x=2, y=2

(c) $\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) x=-2, y=-7

Step-by-step explanation:

Cramer's Rule

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

$\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.$

We call the determinant of the system

$\Delta=\begin{vmatrix}a &b \\c &d \end{vmatrix}$

We also define:

$\Delta_x=\begin{vmatrix}p &b \\q &d \end{vmatrix}$

And

$\Delta_y=\begin{vmatrix}a &p \\c &q \end{vmatrix}$

The solution for x and y is

$\displaystyle x=\frac{\Delta_x}{\Delta}$

$\displaystyle y=\frac{\Delta_y}{\Delta}$

(a) The system to solve is

$\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.$

Calculating:

$\Delta=\begin{vmatrix}1 &1 \\1 &-2 \end{vmatrix}=-2-1=-3$

$\Delta_x=\begin{vmatrix}1 &1 \\4 &-2 \end{vmatrix}=-2-4=-6$

$\Delta_y=\begin{vmatrix}1 &1 \\1 &4 \end{vmatrix}=4-3=3$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1$

The solution is x=2, y=-1

(b) The system to solve is

$\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.$

Calculating:

$\Delta=\begin{vmatrix}4 &-1 \\1 &-1 \end{vmatrix}=-4+1=-3$

$\Delta_x=\begin{vmatrix}6 &-1 \\0 &-1 \end{vmatrix}=-6-0=-6$

$\Delta_y=\begin{vmatrix}4 &6 \\1 &0 \end{vmatrix}=0-6=-6$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2$

The solution is x=2, y=2

$\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.$

Calculating:

$\Delta=\begin{vmatrix}-1 &2 \\1 &2 \end{vmatrix}=-2-2=-4$

$\Delta_x=\begin{vmatrix}0 &2 \\5 &2 \end{vmatrix}=0-10=-10$

$\Delta_y=\begin{vmatrix}-1 &0 \\1 &5 \end{vmatrix}=-5-0=-5$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-10}{-4}=\frac{5}{2}$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-5}{-4}=\frac{5}{4}$

The solution is

$\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

(d) The system to solve is

$\displaystyle \left \{ {{6x-y=-5} \atop {4x-2y=6}} \right.$

Calculating:

$\Delta=\begin{vmatrix}6 &-1 \\4 &-2 \end{vmatrix}=-12+4=-8$

$\Delta_x=\begin{vmatrix}-5 &-1 \\6 &-2 \end{vmatrix}=10+6=16$

$\Delta_y=\begin{vmatrix}6 &-5 \\4 &6 \end{vmatrix}=36+20=56$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{16}{-8}=-2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7$

The solution is x=-2, y=-7

4 0

3 years ago

PLZ I REALLY NEED HELP/ i'll give brainlist and 100 points PLZZZZZ

andrew11 [14]

Answer:

Question 1: Answer would be A.

Question 2: Answer would be B.

Question 3: I cannot answer this one because an image is not provided. I do not have the image of the original triangle LMN. If you message me with a photo of this question, I will be able to answer.

Question 4: This answer may be wrong, but I believe that it would be A. I don't know how to calculate dilation transformations, but after graphing out all the points, I found that (6,4) would be the correct point because the distance between Point S and (6,4) is even and is able to be divided by two.

Question 5: Again, I cannot answer this one because an image is provided. Message me with a photo, I'll answer.

Question 6: Answer would be C.

Question 7: Again, I cannot answer this one because an image is provided. Message me with a photo, I'll answer.

Question 8: Answer would be C.

Step-by-step explanation:

Question 1 Explanation: Answer would be A because the quadrilateral was rotated 180 degrees and dilated to be larger.

Question 2 Explanation: Quadrant IV is on the bottom right, Quadrant I is on the top right. If a triangle were to be rotated from the origin (0,0), it would have to rotate 270 degrees in order to reach Quadrant I.

Question 3 Explanation:

Question 4 Explanation:

Question 5 Explanation:

Question 6 Explanation: Using a rotation calculator, I input the degree of rotation as -90.

Question 7 Explanation:

Question 8 Explanation: Using a rotation calculator, I input the degree of rotation as -180.

7 0

3 years ago