The linear equations are y - 25 = 0.89(x - 20) and y = 0.89x + 7.2
<h3>The slope of the line</h3>
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The two points from the graph are (20, 25) and (38, 41)
The slope of the line is calculated using
m = (y2 - y1)/(x2 - x1)
Substitute the known values in the above equation
m = (41 - 25)/(38 - 20)
Evaluate
m =0.89
<h3>The linear equation in point slope form</h3>
This is calculated as:
y - y1 = m(x - x1)
Substitute the known values in the above equation
y - 25 = 0.89 * (x - 20)
Evaluate
y - 25 = 0.89(x - 20)
<h3>The linear equation in slope-intercept form</h3>
We have:
y - 25 = 0.89(x - 20)
Expand
y - 25 = 0.89x - 17.8
Add 25 to both sides
y = 0.89x + 7.2
Hence, the linear equations are y - 25 = 0.89(x - 20) and y = 0.89x + 7.2
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Answer:
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Step-by-step explanation:
Find the nth term of this number sequence.
2 8 14 20 26
Again write the numbers 1 to 5 above the numbers in the sequence, and leave a spare line again.
n 1 2 3 4 5 (1st row)
(2nd row)
2 8 14 20 26 (3rd row)
Since the sequence is going up by 6, write down your multiples of 6 on the 2nd row.
n 1 2 3 4 5 (1st row)
6n 6 12 18 24 30 (2nd row)
2 8 14 20 26 (3rd row)
Now, to get the numbers in the 3rd row from the 2nd row take off 4.
So, to get from the position numbers (n) to the numbers in the sequence you have to times the position numbers by 6 and take off 4.
Therefore, the nth term = 6n – 4
Given is the function for number of adults who visit fair at day 'd' after its opening, a(d) = −0.3d² + 4d + 9.
Given is the function for number of children who visit fair at day 'd' after its opening, c(d) = −0.2d² + 5d + 11.
Any function f(d) to find excess of children more than adults can be written as follows :-
f(d) = c(d) - a(d).
⇒ f(d) = (−0.2d² + 5d + 11) - (−0.3d² + 4d + 9)
⇒ f(d) = -0.2d² + 0.3d² + 5d - 4d + 11 - 9
⇒ f(d) = 0.1d² + d + 2
$2.26 pounds of horse feed
Answer:
This is the alternate exterior angle theorem
Step-by-step explanation:
This is because angle 2 and angle 7 are in the <u>outer</u> part of each side of opposite lines
