Answer: Tom will need 3 cups of raisins
Step-by-step explanation:
Answer:
see below
Step-by-step explanation:
a(t)=t^(1/2)−t^(−1/2)
We integrate to find the velocity
v(t) = integral t^(1/2)−t^(−1/2) dt
= t ^ (1/2 +1) t ^ (-1/2 +1)
------------ - ----------------- + c where c is the constant of integration
3/2 1/2
v(t) = 2/3 t^ 3/2 - 2 t^ 1/2 +c
We find c by letting t=0 since we know the velocity is 4/3 when t=0
v(0) = 2/3 0^ 3/2 - 2 0^ 1/2 +c = 4/3
0+c =4/3
c = 4/3
v(t) = 2/3 t^ 3/2 - 2 t^ 1/2 +4/3
To find the position function we need to integrate the velocity
p(t) = integral 2/3 t^ 3/2 - 2 t^ 1/2 +4/3 dt
2/3 t ^ (3/2 +1) 2 t ^ (1/2 +1) 4/3t
------------ - ----------------- + ------------- + c
5/2 3/2 1
p(t) = 4/15 t^ 5/2 - 4/3t ^ 3/2 + 4/3t +c
We find c by letting t=0 since we know the position is -4/15 when t=0
p(0) = 4/15 0^ 5/2 - 4/3 0 ^ 3/2 + 4/3*0 +c = -4/15
0 +c = -4/15
c = -4/15
p(t) = 4/15 t^ 5/2 - 4/3t ^ 3/2 + 4/3t -4/15
When you say comparing do you mean dividing or just comparing
The correct answer ic C)12
Answer:
The points for the given to linear equations is (5 , - 2) and (5 , - 1)
The points is plotted on the graph shown .
Step-by-step explanation:
Given as :
The two linear equation are
y = 
x - 1 ...........1
y = 
x - 6 ...........2
Now, Solving both the linear equations
Put the value of y from eq 2 into eq 1
I.e x - 6 = x - 1
Or, x + 
x = 6 - 1
Or, 
x = 5
or, 
x = 5
∴ x = 5
Now, Put the value of x in eq 1
So, y = x - 1
Or, y = × 5 - 1
or, y = 
- 1
Or, y = - 1 - 1
I.e y = -2
So, For x = 5 , y = - 2
Point is (
, 
) = (5 , - 2)
Again , put the value of x in eq 2
So, y = x - 6
Or, y = × 5 - 6
Or, y = 
- 6
Or, y = 4 - 6
I.e y = - 2
So, For x = 5 , y = - 2
Point is (
, 
) = (5 , - 2)
Hence, The points for the given to linear equations is (5 , - 2) and (5 , - 2)
The points is plotted on the graph shown . Answer
