Answer:
- 1) y = 13.5x + 1
- 2) y = 12x + 4
- 3) Sam won the race
Step-by-step explanation:
<h3>Part 1</h3>
Sam's car is 1 ft in front of the start line and its speed is 13.5 ft/s.
<u>The distance after x seconds is:</u>
<h3>Part 2</h3>
Alice's car the speed 12 ft/s and after 3 seconds is 40 ft in front of the start line.
<u>The distance after x seconds is:</u>
- y = 12(x - 3) + 40 = 12x - 36 + 40 = 12x + 4
<h3>Part 3</h3>
<u>After 15 seconds the distance from the start line is:</u>
- Sam ⇒ y = 13.5*15 + 1 = 203.5 ft
- Alice ⇒ y = 12*15 + 4 = 184 ft
As we see Sam is further from the start line than Alice
All you do is subtract 3.05 and 2.74 then do the same thing with the other ones and see what number is between all of them. What grade are you in
Answer:
The pair (0,3) is not a solution to the equation
Step-by-step explanation:
This can be proved by simply replacing the x and y variables in the equation by the x and y values of the pair, and checking if the equation renders a true statement:
By replacing x and y with their values in the pair (0,3), that is x=0 and y=3, in the equation y = 5 - 2x we get:
3 = 5 - 2 (0)
3 = 5 - 0
3 = 5
which is NOT a true statement.
On the other hand, the other two pairs (2,1) and (1,3) render true statements:
1 = 5 - 2 (2)
1 = 5 - 4
1 = 1
and
3 = 5 - 2 (1)
3 = 5 - 2
3 = 3
Answer:
27 degrees
Step-by-step explanation:
First, we can draw this out. Since the altitude is 8000, we can make that the height, and it goes 16000 feet through the air, that can be the length. I have drawn this out, as shown in the image. Note that when drawing, because the airplane is going up and forward, that path should represent the hypotenuse. The angle of elevation is the angle connecting the length and the path, as shown by the angle x in the picture.
We know than tan(x) = opposite/adjacent, so tan(x) here = 8000/16000 = 1/2. Then, arctan(1/2) =x = 27 degrees
Answer:
The required specific heat is 196.94 joule per kg per °C
Step-by-step explanation:
Given as :
The heat generated = Q = 85.87 J
Mass of substance (m)= 34.8 gram = 0.0348 kg
Change in temperature = T2 - T1 = 34.29°C - 21.76°C = 12.53°C
Let the specific heat = S
Now we know that
Heat = Mass × specific heat × change in temperature
Or, Q = msΔt
Or, 85.87 = (0.0348 kg ) × S × 12.53°C
Or , 85.87 = 0.4360 × S
Or, S = 
∴ S = 196.94 joule per kg per °C
Hence the required specific heat is 196.94 joule per kg per °C Answer