y=mx+b is the equation of a line;
m=slope , b= y-intercept
You can find the slope with this following equation: (y(2)-y(1))/(x(2)-x(1))
In this case the points are (0,4) and (-2,-3). The first set being (0,4) and the second (-2,-3). This means (0,4) can be expressed as (x(1),y(1)) and (-2,-3) expressed as (x(2),y(2)). Plugging these numbers into the slope equation gives us: (-3-4)/(-2-0) = -7/-2 = 7/2.
m= 7/2 ; so we have : y= (7/2)x+b
We are give a set of points which it passes through, we can simply plug them in:
4 = (7/2)(0)+b (0 is the x and 4 is the y)
We get 4 = 0 +b .... 4=b
our final equation is : y=(7/2)x+4
Answer:
the answer is D
Step-by-step explanation:
Answer:
A) (1 s, 2.3 s)
B) (-4 m/s², 3.8 m/s²)
Step-by-step explanation:
The car's position which is the distance is given by the equation;
s(t) = t³ - 5t² + 7t
A) Velocity is the first derivative of the distance. Thus;
v(t) = ds/dt = 3t² - 10t + 7
At v = 0, we have;
3t² - 10t + 7 = 0
Using quadratic formula, we have;
t = 1 and t = 2.3
Thus, time at velocity of 0 is t = (1 s, 2.3 s)
B) acceleration is the derivative of the velocity. Thus;
a(t) = dV/dt = 6t - 10
At velocity of 0, we got t = 1 and t = 2.3
Thus;
a(1) = 6(1) - 10 = -4 m/s²
a(2.3) = 6(2.3) - 10 = 3.8 m/s
Thus, a(t) at v = 0 gives; (-4 m/s², 3.8 m/s²)
There is no math :(
so therefore i can not complete the math problem
