The <em>speed</em> intervals such that the mileage of the vehicle described is 20 miles per gallon or less are: v ∈ [10 mi/h, 20 mi/h] ∪ [50 mi/h, 75 mi/h]
<h3>How to determine the range of speed associate to desired gas mileages</h3>
In this question we have a <em>quadratic</em> function of the <em>gas</em> mileage (g), in miles per gallon, in terms of the <em>vehicle</em> speed (v), in miles per hour. Based on the information given in the statement we must solve for v the following <em>quadratic</em> function:
g = 10 + 0.7 · v - 0.01 · v² (1)
An effective approach consists in using a <em>graphing</em> tool, in which a <em>horizontal</em> line (g = 20) is applied on the <em>maximum desired</em> mileage such that we can determine the <em>speed</em> intervals. The <em>speed</em> intervals such that the mileage of the vehicle is 20 miles per gallon or less are: v ∈ [10 mi/h, 20 mi/h] ∪ [50 mi/h, 75 mi/h].
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Answer:
y = 2cos5x-9/5sin5x
Step-by-step explanation:
Given the solution to the differential equation y'' + 25y = 0 to be
y = c1 cos(5x) + c2 sin(5x). In order to find the solution to the differential equation given the boundary conditions y(0) = 1, y'(π) = 9, we need to first get the constant c1 and c2 and substitute the values back into the original solution.
According to the boundary condition y(0) = 2, it means when x = 0, y = 2
On substituting;
2 = c1cos(5(0)) + c2sin(5(0))
2 = c1cos0+c2sin0
2 = c1 + 0
c1 = 2
Substituting the other boundary condition y'(π) = 9, to do that we need to first get the first differential of y(x) i.e y'(x). Given
y(x) = c1cos5x + c2sin5x
y'(x) = -5c1sin5x + 5c2cos5x
If y'(π) = 9, this means when x = π, y'(x) = 9
On substituting;
9 = -5c1sin5π + 5c2cos5π
9 = -5c1(0) + 5c2(-1)
9 = 0-5c2
-5c2 = 9
c2 = -9/5
Substituting c1 = 2 and c2 = -9/5 into the solution to the general differential equation
y = c1 cos(5x) + c2 sin(5x) will give
y = 2cos5x-9/5sin5x
The final expression gives the required solution to the differential equation.
Answer:
Jenny save 4 times 
for the month
Step-by-step explanation:
Given Jenny has saved 
over the last 12 months.
She saved either 
or 
a month.
Let the number of months to save be x
Let the number of months to save be y
Total period is 12 month.
Jenny has saved $3,800 over the last 12 months.
Dividing both side from 50 we get,
Now Multiplying equation 1 by 5 we get,
Now Subtracting Equation 3 by equation 2 we get,
Substituting the value of y in equation 1 we get,
∴ Jenny save 4 Months of and 8 Months of 
in a 12 month duration
R: (-inf, 0]
D: all real #'s
Answer:
in 9 different ways.
Step-by-step explanation:
divide 18 by 2 and u would get the answer
