Greetings from Brasil...
Let's apply the given formula:
A = (1/2)·B·H
The base of this polygon (in this case, the triangle) is B
B = X² - 2X + 6
The height of this polygon is H and is H
H = X + 4
Applying these values (B and H) in the given formula.....
A = (1/2)·B·H
A = (1/2)·(X² - 2X + 6)·(X + 4)
A = (1/2)·(X³ + 2X² - 2X + 24)
A = (X³/2) + X² - X + 12
OR
A = (X³ + 2X² - 2X + 24)/2
Add the three angles and set them to 180
4x-13+15+x+18=180
5x+20=180
5x=160
x=32
Then plug in to get A and C
A=4*32-13
A=115
C=32+18
C=50
6m+1=-23. Do the inverse operation of adding 1 which is to subtract 1 to both sides =====> 6m+1-1= -23-1
You divide by 6 both sides
6m/6=-24/6
m= -4
Brainliest?
Answer:
−35.713332 ; 0.313332
Step-by-step explanation:
Given that:
Sample size, n1 = 11
Sample mean, x1 = 79
Standard deviation, s1 = 18.25
Sample size, n2 = 18
Sample mean, x2 = 96.70
Standard deviation, s2 = 20.25
df = n1 + n2 - 2 ; 11 + 18 - 2 = 27
Tcritical = T0.01, 27 = 2.473
S = sqrt[(s1²/n1) + (s2²/n2)]
S = sqrt[(18.25^2 / 11) + (20.25^2 / 18)]
S = 7.284
(μ1 - μ2) = (x1 - x2) ± Tcritical * S
(μ1 - μ2) = (79 - 96.70) ± 2.473*7.284
(μ1 - μ2) = - 17.7 ± 18.013332
-17.7 - 18.013332 ; - 17.7 + 18.013332
−35.713332 ; 0.313332
