Let
T--------> the total price of the vehicle
x--------> amount paid
y--------> amount owed
we know that
the total cost of the vehicle is equal to the amount paid plus the amount owed
so
Find the amount paid
Find the amount owed
Find the total cost of the vehicle
therefore
the answer is
the total price of the vehicle is
The total price of the vehicle is 28,466
To determine the total price of the vehicle, we can solve this question using the area model (please refer to the attachment below)
Firstly, let’s use the area model method to calculate the total price of the vehicle.
Therefore, we have:
567 x 48
= (500 + 60 + 7) x (40 + 8)
= (500 x 40) + (500 x 8) + (60 x 40) + (60 x 8) + (7 x 40) + (7 x 8)
= 20000 + 4000 + 2400 + 480 + 280 + 56
= 27216.
Recall, Jaycee still owes $1250 after 48 months; therefore we can derive the total price of the vehicle by adding the total money for the whole 48 months and the amount Jayson owes.
Therefore we have:
27216 + 1250
= 28466
Thus, the total price of the vehicle is $28,466
The area model refers to a rectangular diagram used in solving multiplication and division problems in mathematics.
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f(x) = –2x2 + 16x – 35
f(x) = x2 – 4x + 5
f(x) = 2x2 – 16x + 35
Answer:
Step-by-step explanation:
In order to solve the question, we have to derivate each function.
1) f(x) = x2 +4x -11
Then,
f'(x)= 2x +4
Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:
f'(4)= 2*4 +4 = 12 ≠ 0 then this function doesn't not have a minimum at (4, -3)
2) f(x) = –2x2 + 16x – 35
Then,
f'(x)= -4x +16
Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:
f'(4)= -4*4 16 = 4 ≠ 0 then this function have a critical point at (4, -3)
then,
f''(4) =-4 <0 then we have a minimum at (4, -3)
3) f(x) = x2 – 4x + 5
Then,
f'(x)= 2x -4
Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:
f'(4)= 2*4 -4 = 4 ≠ 0 then this function doesn't not have a minimum at (4, -3)
4) f(x) = 2x2 – 16x + 35
Then,
f'(x)= 4x -16
Now, if f'(4)=0 we would have a critical point, which could be a minimum. Let's find out:
f'(4)= 4*4 -16 = 4 ≠ 0 then this function doesn't not have a minimum at (4, -3)