Answer: 1/40 or .025
See the attached image for the explanation.
Hope this helps! Please make me the brainliest, it’s not necessary but appreciated, I put a lot of effort and research into my answers. Have a good day, stay safe and stay healthy.
Perpendicular lines refers to a pair of straight lines that intercept each other. The slopes of this lines are opposite reciprocal, meaning that it's multiplication is -1.
On this case they give you the equation of a line and a point, and is asked to find the equation of a line that is perpendicular to the given one, and that passes through this point.
-2x+3y=-6 Add 2x in both sides
3y=2x-6 Divide by 3 in both sides to isolate y
y=2/3x-6/3
The slope of the given line is 2/3, which means that the slope of a line perpendicular to this one, needs to be -3/2. Now you need to find the value of b or the y-intercept by substituting the given point into the formula y=mx+b, where letter m represents the slope.
y=mx+b Substitute the given point and the previous slope found
-2=(-3/2)(6)+b Combine like terms
-2=-9+b Add 9 in both sides to isolate b
7=b
The equation that represents the line perpendicular to -2x+3y=-6 and that passes through the point (6,-2), is y=-3/2x+7.
Hey!
You have to use the Pythagorean Theorem
a² + b² = c²
a = 10
10² = 10 × 10 = 100
b = 11
11² = 11 × 11 = 121
100 + 121 = 221
√221 = 14.9(rounded)
Answer ⇒ 14.9
Hope this helps! :)
The blank answers are 340 and 0.34
Answer: The average was 9 years old find the average age of both groups is 10 years old.
Step-by-step explanation:
Formula foe average : 
Given : The first group of students consists of 10 and their average age was 13 years old.
i.e. 
(1)
The next group consisted of 30 students and their average was 9 years old.
i.e. 
(2)
Then from (1) and (2) , the sum of both groups (first group and next group )students = 130+270 =400
Combined students of both groups (first and next group )= 10+30=40
Now , the average of both groups =
Hence, the average was 9 years old find the average age of both groups is 10 years old.
