4,10 I am only in 5th grade. Ordered pairs might be what ur working with
Answer:
2(d-vt)=-at^2
a=2(d-vt)/t^2
at^2=2(d-vt)
Step-by-step explanation:
Arrange the equations in the correct sequence to rewrite the formula for displacement, d = vt—1/2at^2 to find a. In the formula, d is
displacement, v is final velocity, a is acceleration, and t is time.
Given the formula for calculating the displacement of a body as shown below;
d=vt - 1/2at^2
Where,
d = displacement
v = final velocity
a = acceleration
t = time
To make acceleration(a), the subject of the formula
Subtract vt from both sides of the equation
d=vt - 1/2at^2
d - vt=vt - vt - 1/2at^2
d - vt= -1/2at^2
2(d - vt) = -at^2
Divide both sides by t^2
2(d - vt) / t^2 = -at^2 / t^2
2(d - vt) / t^2 = -a
a= -2(d - vt) / t^2
a=2(vt - d) / t^2
2(vt-d)=at^2
Answer:
<h2>y=2x+0 (y=2x)</h2>
Step-by-step explanation:
so we know that y=mx+c
c is the point where the line crosses the y-axis
y=mx+0
m is the gradient
the gradient is rise/run
pick any two points
so it is 2/1
2/1 is 2
therefore it is
y=2x+0
Answer:
15/32
10/9 = 1 1/9
26/15 = 1 11/15
Step-by-step explanation:
3/8 / 4/5 = 3/8 * 5/4 = 15/32
8/9 / 4/5 = 8/9 * 5/4 = 40/36 = 10/9 = 1 1/9
2 3/5 / 1 1/2 = 13/5 / 3/2 = 13/5 * 2/3 = 26/15 = 1 11/15
