The system of equation that represents this situation is y = 20x + 15 for day and night kernel and y = 30x + 5 for bark time hotel.
A linear equation is in the form:
y = mx + b
where y, x are variables, m is the rate of change and b is the initial value of y.
Let y represent the total charges for for boarding a pet in x days.
Day and night kennel charges $20 per day plus a one-time food fee of $15 to board a pet. Hence:
y = 20x + 15
Bark time hotel charges $30 per day plus a one-time food fee of $5. Hence:
y = 30x + 5
The system of equation that represents this situation is y = 20x + 15 for day and night kernel and y = 30x + 5 for bark time hotel.
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The equation
shows that the diagonals are congruent perpendicular bisectors.
The vertices of the square are given as:
- c = (1,1)
- d = (3,1)
- e =(3,-1)
- f = (1,-1)
<h3>How to determine the
congruent perpendicular bisectors.</h3>
Start by calculating the slope of diagonal ce using:

So, we have:



Next, calculate the slope of diagonal df using:

So, we have:



The slopes of both diagonals are:


By comparing both slopes, we have:

i.e.

Hence,
shows that the diagonals are congruent perpendicular bisectors.
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Cos(2θ) = cos²θ - sin²θ = (1-sin²θ)-sin²θ = 1-2(7/9)^2 = -17/81
The number of ways the skaters can finish the competition is 40,320 ways
The different ways 3 of the skaters finish first, second and third is 56 ways
Given the following
- Number of skaters featured = 8 skaters
If the skaters finish the competition, the number of different ways the skaters finish the competition is expressed as:
8! = 8*7*6*5*4*3*2
8! = 56*30*24
8! = 40,320.
The number of ways the skaters can finish the competition is 40,320 ways
If 3 of the skaters finish first, second and third, the number of ways this can be done is given as:
8C3 = 8!/(8-3)!3!
8C3 = 8!/5!3!
8C3 = 8*7*6*5!/5!3!
8C3 = 56 ways
Hence the different ways 3 of the skaters finish first, second and third is 56 ways
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