Given the position function
<em>s(t)</em> = 3 - cos(<em>t</em> )
differentiate once to get the velocity, and twice to get the acceleration:
<em>v(t)</em> = <em>s'(t)</em> = sin(<em>t </em>) → <em>v</em> (4.3 s) ≈ -0.92
<em>a(t)</em> = <em>s''(t)</em> = <em>v'(t)</em> = cos(<em>t </em>) → <em>a</em> (4.3 s) ≈ -0.40
The acceleration is negative, so the particle is slowing down.
14.99*.12=1.80
14.99-1.8=13.19
13.19*.0625=.94
13.19+.94=14.13
final price 14.13
Answer:
b = -6
Step-by-step explanation:
We use the general equation a*x^2 + bx + c = 0, the Quadratic Equation.
Compare it to 2x^2 - 6x - 20 = 0
b = -6 here
I'm not sure what you're supposed to round to, but since this is money, I would suggest including the dollar sign in front of the answers and including round to the hundredth's decimal place. Otherwise, you're answers are correct.
Answer:
Option B is correct.
Step-by-step explanation:
4x+3y-z= -6 eq(1)
6x-y+3z= 12 eq(2)
8x+2y+4z=6 eq(3)
We need to solve these equations and find the value of x, y and z.
Multiply equation 2 with 3 and then add equation 1 and 2
18x -3y +9z = 36
4x +3y -z = -6
_____________
22x + 8z = 30 eq(3)
Multiply equation 2 with 2 and add with equation 3
12x -2y + 6z = 24
8x +2y +4z = 6
____________
20x + 10 z = 30 eq(4)
Multiply equation 3 with 10 and equation 4 with 8 and then subtract
220x + 80z = 300
160x + 80z = 240
- - -
_________________
60x = 60
x= 60/60
x= 1
Putting value of x in equation 3
22x + 8z = 30
22(1) + 8z = 30
8z = 30 - 22
8z = 8
z = 8/8
z=1
Putting value of x and z in equation 1
4(1)+3y-(1)=-6
4 + 3y -1 = -6
3 + 3y = -6
3y = -6 -3
3y = -9
y = -3
so, Option B x=1, y=3 and z=1 is correct
