2/14 just add the numerators and denomentators
Answer:
note :
The equation of a linear function in point-slope form is y – y1 = m(x – x1)
The point is A (x1 , y1)
Step-by-step explanation:
in this exercice : y - 4 = 5 +(1/2)x
y - 4 = 0.5 ( x - 5/0.5)
y - 4 = 0.5( x - 10).....
note : the standard forme is : y = ax +b
in this exercice : y = (1/2)x+9
The answer is y=4x-4. Have a nice day!
Answer:
The entering car is going to catch up with the other car after 20.76 seconds.
Step-by-step explanation:
The car leaving the pit area is moving with a constant accelaration, then we can calculate the speed it has when entering the main speedway using the following equation:

Where
is the initial speed (which is zero given that the car starts froms rest),
is the car's accelaration and
is the time passed until it reaches the main speedway.


For an object travelling at a constant accelation, the displacement
at a time
can be calculate using the following equation:
<em>(equation 1)</em>
Let's consider
the point where the first car enters the main speedway. If
is the point where this car catch up with other one after
seconds, we may rewrite the <em>equation 1</em> like this:

<em>(equation 2)</em>
In other hand, the second car is travelling at a constant speed
when it meets the entering car. Then, its displacement
after
seconds can be calculated using the following formula:
<em>(equation 3)</em>
The entering car is going to catch up with the other one when
, so when can find how much time this will require equaling <em>equations 2 and 3</em> and isolating 


![[\frac{1}{2}(a*t_2)-(v-s_1)]*t_2=0](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1%7D%7B2%7D%28a%2At_2%29-%28v-s_1%29%5D%2At_2%3D0)



So, the entering car is going to catch up with the other car after 20.76 seconds.