The equation of the perpendicular line to the given line is: y = -5/4x - 30.
<h3>What is the Equation of Perpendicular Lines?</h3>
The slope values of two perpendicular lines are negative reciprocal of each other.
Given that the line is perpendicular to y = 4/5x+23, the slope of y = 4/5x+23 is 4/5. Negative reciprocal of 4/5 is -5/4.
Therefore, the line that is perpendicular to it would have a slope (m) of -5/4.
Plug in m = -5/4 and (x, y) = (-40, 20) into y = mx + b to find b:
20 = -5/4(-40) + b
20 = 50 + b
20 - 50 = b
b = -30
Substitute m = -5/4 and b = -30 into y = mx + b:
y = -5/4x - 30
The equation of the perpendicular line is: y = -5/4x - 30.
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Answer:
(a) y(x)=53+7x
(b) 179
Step-by-step explanation:
Since the first row has 60 seats and next row has 7 additional seats then we can represent it as
First row=60
Second row=60+7=67
Third row=67+7=74
The difference is always 7. If you deduct 7 from dirst row we get 60-7=53 seats
To get rhe number of seats in any row x then let y be the number of seats in row x
y=53+7(x)
For raw 1
Y=53+7(1)=60
For raw 2
Y=53+7(2)=67
Therefore, the formula for number of seats at any row will be
y(x)=53+7(x)
(b)
Using the above formula
y(x)=53+7(x)
Replace x with 18 hence
Y(18)=53+7*(18)=179 seats
Answer:
11 ;
Step-by-step explanation:
Given the data:
30 27 22 25 24 25 24 15 35 35 33 52 49 10 27 18 20 23 24 25 30 24 24 24 18 20 25 27 24 32 13 13 21 2 37 35 32 33 29 3 28 28 25 29 31
Number of classes = 5
Class width : Range / number of classes
Class width = (maximum - minimum) / 5
Class width = (52 - 2) / 5 = 50/5 = 10 + 1 = 11
For the frequency table showing class limits, class boundaries, midpoints, frequencies, relative frequencies, and cumulative frequencies and histogram
Kindly check attached picture
Answer:
26.4
31.2
33.3
These are the easy questions, the rest will have a square root and a a number outside.
