Answer:
the first one on the left closest to the question.
Step-by-step explanation:
Graph the inequality by finding the boundary line, then shading the appropriate area.
The longest possible altitude of the third altitude (if it is a positive integer) is 83.
According to statement
Let h is the length of third altitude
Let a, b, and c be the sides corresponding to the altitudes of length 12, 14, and h.
From Area of triangle
A = 1/2*B*H
Substitute the values in it
A = 1/2*a*12
a = 2A / 12 -(1)
Then
A = 1/2*b*14
b = 2A / 14 -(2)
Then
A = 1/2*c*h
c = 2A / h -(3)
Now, we will use the triangle inequalities:
2A/12 < 2A/14 + 2A/h
Solve it and get
h<84
2A/14 < 2A/12 + 2A/h
Solve it and get
h > -84
2A/h < 2A/12 + 2A/14
Solve it and get
h > 6.46
From all the three inequalities we get:
6.46<h<84
So, the longest possible altitude of the third altitude (if it is a positive integer) is 83.
Learn more about TRIANGLE here brainly.com/question/2217700
#SPJ4
rewrite -1 1/3 as a decimal: -1.5
-2.1 - x = -1.5
Subtract -2.1 From both sides:
X = -1.5 - -2.1 = -1.5 + 2.1
X = 0.6
Answer: B. 0.6
Answer:
17
Step-by-step explanation:
Using the law of cosines in ΔABC
b² = a² + c² - (2ac cosB) ← substitute in values
= 25² + 28² - (2 × 25 × 28 × cos36.9°)
= 625 + 784 - 1400 cos36.9°
= 1409 - 1400 cos36.9° ( Take the square root of both sides )
b = 
≈ 17 ( nearest whole number )
Answer:
hi sh Sheet sh I monorailg Jericho improve Odom Ybor
