To find the gradient you do change in y/change in x so...
-12 - 0
————- = -12/4 = -3
0 - - 4
Hope this helps! Any questions let me know :)
Answer:
Step-by-step explanation:
Given that acceleration of an object is
is the solution to the differential equation
Since v(0) =7
we get ln 7 = C
Hence 
since velocity is rate of change of distance s we have
![v=\frac{ds}{dt} =7e^{-2t}\\s= [tex]s(t) =\frac{-7}{2} (e^{-2t})+C)[](https://tex.z-dn.net/?f=v%3D%5Cfrac%7Bds%7D%7Bdt%7D%20%3D7e%5E%7B-2t%7D%5C%5Cs%3D%20%5Btex%5Ds%28t%29%20%3D%5Cfrac%7B-7%7D%7B2%7D%20%28e%5E%7B-2t%7D%29%2BC%29%5B)
substitute t=0 and s=0
So solution for distance is
Answer:
f(4) = 49
f(-3) = -35
Step-by-step explanation:
f(4) = 12(4) + 1 = 48 + 1 = 49
f(-3) = 12(-3) + 1 = -36 + 1 = -35
Answer:
y = 18 and x = -2
Step-by-step explanation:
y = x^2+bx+c To find the turning point, or vertex, of this parabola, we need to work out the values of the coefficients b and c. We are given two different solutions of the equation. First, (2, 0). Second, (0, -14). So we have a value (-14) for c. We can substitute that into our first equation to find b. We can now plug in our values for b and c into the equation to get its standard form. To find the vertex, we can convert this equation to vertex form by completing the square. Thus, the vertex is (4.5, –6.25). We can confirm the solution graphically Plugging in (2,0) :
y=x2+bx+c
0=(2)^2+b(2)+c
y=4+2b+c
-2b=4+c
b=-2+2c
Plugging in (0,−14) :
y=x2+bx+c
−14=(0)2+b(0)+c
−16=0+b+c
b=16−c
Now that we have two equations isolated for b , we can simply use substitution and solve for c . y=x2+bx+c 16 + 2 = y y = 18 and x = -2
Answer:
The answer is 24
Step-by-step explanation:
24/4 is 6 and then you add 6 to 3 which is 9
