G^−m ÷ g^n
1st g^−m=1/g^m, hence g^−m ÷ g^n = (1/g^m) /(g^n)==> 1/(g^m)(g^(n)
==> 1/(g^m+n) or g^(-m-n)
Answer:
Step-by-step explanation:
3(9x - 14/3x + 4 - 1 ≥ 2)
27x - 14x + 12 - 3 ≥ 6
13x + 9 ≥ 6
13x ≥ -3
x ≥ -3/13
Answer: b. (6,27)
Step-by-step explanation:
y=5x- 3 x=6<---- put in the equation
y=5(6)-3 <------- mulitply 5 and 6
y=30-3 <---- subtracted
y= 27. <---- y value.
Answer being (6,27)
Hope it helps :)
Answer:
you subtract then you add on how much money you get know
Step-by-step explanation:
hope that helps
Answer:
a) The differential equation for the velocity is given by
(dv/dt) = k(250 - v)
b) v(t) = 250 - e⁽⁵•⁵² ⁻ ᵏᵗ⁾
With units of km/h
Step-by-step explanation:
Acceleration, a ∝ (250 - v)
But acceleration is widely given as dv/dt
(dv/dt) ∝ (250 - v)
(dv/dt) = k(250 - v)
where k = constant of proportionality
(dv/dt) = k(250 - v)
b) (dv/dt) = k(250 - v)
dv/(250 - v) = k dt
∫ dv/(250 - v) = ∫ k dt
- In (250 - v) = kt + c (where c is the constant of integration)
v(0) = 0; meaning, at t = 0, v = 0
- In 250 = 0 + C
c = - In 250 = - 5.52
- In (250 - v) = kt - 5.52
In (250 - v) = 5.52 - kt
250 - v = e⁽⁵•⁵² ⁻ ᵏᵗ⁾
v = 250 - e⁽⁵•⁵² ⁻ ᵏᵗ⁾
With units of km/h
i hope this work for you
and sory if im wrang
