The area is 14! hope this helps
After a little manipulation, the given diff'l equation will look like this:
e^y * dy = (2x + 1) * dx.
x^2
Integrating both sides, we get e^y = 2------- + x + c, or e^y = x^2 + x + c
2
Now let x=0 and y = 1, o find c:
e^1 = 0^2 + 0 + c. So, c = e, and the solution is e^y = x^2 + x + e.
The distance travelled is 10 m
The velocity gained at the end of the time is 2 m/s
<h3>Motion</h3>
From the question, we are to determine distance travelled and the velocity gained
From one of the equations of motion for <u>linear motion</u>, we have that
S = ut + 1/2at²
Where S is the distance
u is the initial velocity
t is the time taken
and a is the acceleration
First, we will calculate the acceleration
Using the formula,
F = ma
Where F is the force
m is the mass
and a is the acceleration
∴ a = F/m
Where F is the force
and a is the acceleration
From the given information,
F = 50 N
m = 250 kg
Putting the parameters into the equation,
a = 50/250
a = 0.2 m/s²
Thus,
From the information,
u = 0 m/s (Since the object was initially at rest)
t = 10 s
S = 0(t) + 1/2(0.2)(10)²
S = 10 m
Hence, the distance travelled is 10 m
For the velocity
Using the formula,
v = u + at
Where v is the velocity
v = 0 + 0.2×10
v = 2 m/s
Hence, the velocity gained at the end of the time is 2 m/s
Learn more on Motion here: brainly.com/question/10962624
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-1 < c + 2 < 3....subtract 2 from all sections
-1 - 2 < c + 2 - 2 < 3 - 2...simplify
-3 < c < 1
==============================
32 > 16 - 4g > 12....subtract 16 from all sections
32 - 16 > 16 - 16 - 4g > 12 - 16....simplify
16 > -4g > -4 ...now divide all sections by -4, and change inequality signs
16/-4 < (-4/-4)g < -4/-4...simplify
-4 < g < 1
============================
6y + 1 > = 10
6y > = 10 - 1
6y > = 9
y > = 9/6 which reduces to 3/2 or 1 1/2
-3/2y > = 9 ....multiply both sides by -2/3, cancelling the -3/2 on the left...and change the inequality sign
y < = 9 * -2/3
y < = -18/3 which reduces to - 6
so y > = 9/6(or 1 1/2) or y < = -6
