Answer:
<h2>14mph</h2>
Step-by-step explanation:
Given the gas mileage for a certain vehicle modeled by the equation m=−0.05x²+3.5x−49 where x is the speed of the vehicle in mph. In order to determine the speed(s) at which the car gets 9 mpg, we will substitute the value of m = 9 into the modeled equation and calculate x as shown;
m = −0.05x²+3.5x−49
when m= 9
9 = −0.05x²+3.5x−49
−0.05x²+3.5x−49 = 9
0.05x²-3.5x+49 = -9
Multiplying through by 100
5x²+350x−4900 = 900
Dividing through by 5;
x²+70x−980 = 180
x²+70x−980 - 180 = 0
x²+70x−1160 = 0
Using the general formula to get x;
a = 1, b = 70, c = -1160
x = -70±√70²-4(1)(-1160)/2
x = -70±√4900+4640)/2
x = -70±(√4900+4640)/2
x = -70±√9540/2
x = -70±97.7/2
x = -70+97.7/2
x = 27.7/2
x = 13.85mph
x ≈ 14 mph
Hence, the speed(s) at which the car gets 9 mpg to the nearest mph is 14mph
Answer:
165m
Question:
Artur, Olga and Wiktor participated in the race. They started from the same place at the same time and run at constant speeds. When Artur finished the race, Olga was 15 m to the finish,
and Wiktor was 35 m. When Olga finished the race, Wiktor remained 22 m to the finish. At what distance was the race held?
Step-by-step explanation:
Let Artur distance covered be =>x
When Artur covered distance x:
Olga was 15m from x
Olga = x - 15
Wiktor was 35m from x
Wiktor = x - 35
When Olga covered distance x:
Wiktor was 22m from x
The ratio of Olga to Wiktor:
(x-15)/(x-35) = x/(x-22)
Cross multiply
(x-15)(x-22) = x(x-35)
x² - 15x - 22x + 330 = x² - 35x
x² -37x +330 - x² + 35x = 0
-2x = -330
x = (-330)/(-2)
x = 165m
The race held at 165m distance
Answered by Mimiwhatsup: If two parallel lines are cut by a third line, the third line is called the transversal.
Relationships:
(1) Corresponding angles
(2) Vertically Opposite angles
(3) Alternate interior angles
(4) Alternate exterior angles
Answer:
a is the answer because I just did that question and got 100 percent.
Answer:
-2.0000 times -2.000 = 4
Step-by-step explanation: