Every year 39 million cars cross the Golden Gate Bridge in San Francisco. Which kind of function best models the relationship be
2 answers:
Answer:
Linear
Step-by-step explanation:
Linear function : if the value increases with the constant value per period,
It can written as y = a + bx
Where, a = initial value, b = constant value per period,
Exponential function : if the value increases with the constant rate per period,
It can be written as,
Where, a = initial value, b = growth factor per period.
∵ Every year 39 million cars cross the Golden Gate Bridge in San Francisco.
Then after x years, the number of cars ( in million )
f(x) = 39x
⇒ f(x) = 0 + 39x
By the above explanation,
Hence, the function that represents the number of cars that have crossed the Golden Gate Bridge is linear.
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The given information are the coordinates of points H, Z and B. So, the first step to do is to plot these points on a Cartesian plane as shown in the attached picture. We can deduce visually that horse Z is closer to the barns than horse H. But, to further justify the answer, we have to provide the magnitude of the distance of horses H and Z to barn B. In this approach, we use the distance formula:
d = √(x₂ - x₁)² + (y₂ - y₁)²
Use the coordinates of the two points to know their linear distances. These are represented by the red and green lines for each of the horses.
Distance between H and B = √(⁻3 - 1)² + (⁻9-10)²
Distance between H and B = √(⁻3 - 1)² + (⁻9-10)²
Distance between H and B = 19.4165
Since the scale is 1 unit = 100 meters, the actual distance between horse H and barn B is 19.416*100 = 1,941.65 meters
Distance between Z and B = √(⁻3 - 10)² + (⁻9-1)²
Distance between Z and B = √(⁻3 - 10)² + (⁻9-1)²
Distance between Z and B = 16.4012
Since the scale is 1 unit = 100 meters, the actual distance between horse Z and barn B is 16.4012*100 = 1,640.12 meters
Comparing the distances: 1,941.65 meters > 1,640.12 meters. Therefore, it justifies that horse Z is nearer to the barn.
d = √(x₂ - x₁)² + (y₂ - y₁)²
Use the coordinates of the two points to know their linear distances. These are represented by the red and green lines for each of the horses.
Distance between H and B = √(⁻3 - 1)² + (⁻9-10)²
Distance between H and B = √(⁻3 - 1)² + (⁻9-10)²
Distance between H and B = 19.4165
Since the scale is 1 unit = 100 meters, the actual distance between horse H and barn B is 19.416*100 = 1,941.65 meters
Distance between Z and B = √(⁻3 - 10)² + (⁻9-1)²
Distance between Z and B = √(⁻3 - 10)² + (⁻9-1)²
Distance between Z and B = 16.4012
Since the scale is 1 unit = 100 meters, the actual distance between horse Z and barn B is 16.4012*100 = 1,640.12 meters
Comparing the distances: 1,941.65 meters > 1,640.12 meters. Therefore, it justifies that horse Z is nearer to the barn.
Solutions
In Matrix we use initially based on systems of linear equations.The matrix method is similar to the method of Elimination as but is a lot cleaner than the elimination method.Solving systems of equations by Matrix Method involves expressing the system of equations in form of a matrix and then reducing that matrix into what is known as Row Echelon Form.<span>
Calculations
</span>⇒ <span>Rewrite the linear equations above as a matrix
</span>
![\left[\begin{array}{ccc}3&5&42\\6&5&54\\\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%265%2642%5C%5C6%265%2654%5C%5C%5Cend%7Barray%7D%5Cright%5D%20)
⇒ Apply to Row₂ : Row₂ - 2 <span>Row₁
</span>
![\left[\begin{array}{ccc}3&5&42\\0&-5&-30\\\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%265%2642%5C%5C0%26-5%26-30%5C%5C%5Cend%7Barray%7D%5Cright%5D%20)
⇒ <span>Simplify rows
</span>
![\left[\begin{array}{ccc}3&5&42\\0&1&6\\\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%265%2642%5C%5C0%261%266%5C%5C%5Cend%7Barray%7D%5Cright%5D%20)
Note: The matrix is now in echelon form.
<span>The steps below are for back substitution.
</span>
⇒ Apply to Row₁<span> : Row</span>₁<span> - </span>5 Row₂
![\left[\begin{array}{ccc}3&0&12\\0&1&6\\\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%260%2612%5C%5C0%261%266%5C%5C%5Cend%7Barray%7D%5Cright%5D%20)
⇒ <span>Simplify rows
</span>
![\left[\begin{array}{ccc}1&0&4\\0&1&6\\\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%264%5C%5C0%261%266%5C%5C%5Cend%7Barray%7D%5Cright%5D%20)
⇒ <span>Therefore,
</span>
<span>
</span>
In Matrix we use initially based on systems of linear equations.The matrix method is similar to the method of Elimination as but is a lot cleaner than the elimination method.Solving systems of equations by Matrix Method involves expressing the system of equations in form of a matrix and then reducing that matrix into what is known as Row Echelon Form.<span>
Calculations
</span>⇒ <span>Rewrite the linear equations above as a matrix
</span>
⇒ Apply to Row₂ : Row₂ - 2 <span>Row₁
</span>
⇒ <span>Simplify rows
</span>
Note: The matrix is now in echelon form.
<span>The steps below are for back substitution.
</span>
⇒ Apply to Row₁<span> : Row</span>₁<span> - </span>5 Row₂
⇒ <span>Simplify rows
</span>
⇒ <span>Therefore,
</span>
<span>
</span>