1.1 Factoring x2-2x-3
The first term is, x2 its coefficient is 1 .
The middle term is, -2x its coefficient is -2 .
The last term, "the constant", is -3
Step-1 : Multiply the coefficient of the first term by the constant 1 • -3 = -3
Step-2 : Find two factors of -3 whose sum equals the coefficient of the middle term, which is -2 .
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -3 and 1
x2 - 3x + 1x - 3
Step-4 : Add up the first 2 terms, pulling out like factors :
x • (x-3)
Add up the last 2 terms, pulling out common factors :
1 • (x-3)
Step-5 : Add up the four terms of step 4 :
(x+1) • (x-3)
Which is the desired factorization
A histogram is a graphical representation of data. A histogram is uniform when it forms almost a straight, horizontal line. It is symmetric when it forms a bell shape, equal parts to both sides. It is skewed when most of the data falls to the left or right.
With reference to the image attached below, a histogram is uniform if it forms a straight line, symmetrical if it forms a bell-shaped curve, and skewed.
When a histogram is uniform, it appears like a straight line is formed. When the histogram is symmetrical, it forms a normal bell-shaped curve as shown in the image attached below.
Also, the image attached below shows how a skewed histogram will appear like.
In summary, a histogram is uniform if it forms a straight line, symmetrical if it forms a bell-shaped curve, and skewed as shown in the image below.
Learn more about skewed, uniform, and symmetrical histogram on:
#SPJ5
-108,-90,-72,-54
Answer:
V =12.167 cm^3
Step-by-step explanation:
V = s^3 is the volume of a cube where s is the side length
The side length of the cube is 2.3 cm
V = ( 2.3 cm) ^3
V =12.167 cm^3
Hey there!
The FORMULA:
V = s^3
Your EQUATION:
v = 2.3^3
SOLVING for the EQUATION
v = 2.3^3
2.3^3 = v
(2.3 cm)^3
≈ 2.3 cm * 2.3 cm * 2.3 cm
≈ 5.29 cm * 2.3 cm
≈ 12.167 cm^3
v ≈ 12.167 cm^3
Therefore, your answer is: 12.167 cm^3
Good luck on your assignment and enjoy your day!
~Amphitrite1040:)