Let us always base on the present age. We denote this by x. Now, the age he had 2 years ago would then be denoted as (x-2). Let's equate this to 20 years old.
x - 2 = 20
x = 20 + 2
x = 22 years old
He should be 22 years old now.
Let's check the other condition. After 1 year, his age should be (x+1). Let's equate this to 23 years old.
x + 1 = 23
x = 23 - 1
x = 22
Thus, this is possible if Reuben is 22 years old as of the present.
3x + 1 = y
2x + 3y = 14
To solve this system of equations, we are going to use the substitution method. Substitution the equation where the variable is isolated into the second equation. In this system of equations, y is isolated, so we will replace y in the second equation with 3x + 1.
2x + 3y = 14
2x + 3(3x + 1) = 14
2x + 9x + 3 = 14
We will add the like terms and subtract 3 from both sides of the equation.
11x + 3 = 14
11x = 11
x = 1
In this system of equations, x is equal to 1. Now we will go back and solve for y, plugging in 1 for x.
3(1) + 1 = y
2(1) + 3y = 14
3 + 1 = y
2 + 3y = 14
4 = y
3y = 2
4 = y
4 = y
The solution to this system of equations is (1, 4).
7.855 mm maybe the answer unless rounded to the tenths place to be 7.9 mm.
Answer:
x=5/2 + a/2, y= 5/2 - a/2
Step-by-step explanation:
We can solve this system using addition of the equations
x+y=5
<u>x-y=a</u>
2x =(5+a)
x=(5+a)/2= 5/2 +a/2
x=5/2 +a/2
x-y=a
y = x - a , x=5/2 + a/2
y= 5/2 + a/2-a
y= 5/2 - a/2
Answer:
Step-by-step explanation:
The position of an object moving horizontally after t seconds is given by the function
s = 3t - t³
a) The object is stationary when there is no external force acting on the body. When the body is at rest, the body remains in a position and here is no distance covered by the object i.e s = 0
b) velocity is the change in displacement of a body with respect to time.
v = ds/dt
S = 3t - t³
V = ds/dt = 3-3t²
at t = 2
Velocity = 3-3(2)²
Velocity = 3-12
Velocity = -9m/s
c) acceleration is the change in velocity of a body with respect to time.
acceleration = dv/dt
If v = 3-3t²
a = dv/dt = -6t
When v = 0
0 = 3-3t²
-3 = -3t²
t² = 1
t = ±√1
t = 1sec
The acceleration of the object at v = 0 occurs at t = 1sec and -1sec
a = -6(1)
a = -6m/s²
d) Given the speed of the body v modelled by the function
v = 3-3t²
The speed is decreasing when it is less than zero as shown:
3-3t²< 0
3<3t²
1<t²
±1<t
1<t and -1<t
t>±1
t >1 and t>-1
The speed is decreasing when
3-3t²<0 and t>1 or t>-1
