41°
65°
74°
The problem asks for a location that is equidistant from towns A and B and lies on the given road. Calculating the midpoint of A and B, we get (5, 1.5). However, this point does not lie on the road denoted by -x + 7y = -4. So, we cannot determine the exact location of the school with the given conditions.
In this problem, the location of the school should be the midpoint of the line between towns A and B as it is equidistant from both towns. First, let's calculate the midpoint (M) coordinates. The formulas for finding the x and y coordinates of the midpoint are (x1 + x2) / 2 and (y1 + y2) / 2 respectively. Using these formulas, we get the coordinates of M as (2+8)/2, (-2+5)/2 = (5, 1.5). However, we should ensure that this point lies on the given road, which is denoted by the equation -x + 7y = -4. Substituting the coordinates of M in the equation, we get -5 + 7*1.5 = -5 + 10.5 = 5.5 which is not equal to -4. So, (5, 1.5) is not a valid location for the school. Unfortunately, with the given conditions, we cannot determine the exact location of the school. Additional information or revision of the conditions might be necessary to solve this problem.
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Sarah is correct because if Ursula is earning $9 per hour then every hour you can add 9. Lets say h=2 then you would multiply 9*2 to get how much she spent. Josh is incorrect because dividing would subtract the amount of money every our.
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b)30t people per year^2
c)600t people per year
d)36t people per year^2
The acceleration in the population t years from 1996 is 60t people per year^2.
A function assigns the value of each element of one set to the other specific element of another set.
It is given that a population grows from an initial size of 10 people to an amount P(t), given by P(t)=10(3+0.5t+t³) where t is measured in years from 1996.
Now, in order to find the acceleration in the population t years from 1996, we need to find the second derivative of the populationfunction, therefore,
Hence, the acceleration in the population t years from 1996 is 60t people per year^2.
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