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kobusy [5.1K]
3 years ago
14

Simplify the expression. 15y^7-6-y^5 + 3y^4 / 6y^4

Mathematics
1 answer:
Vinil7 [7]3 years ago
3 0

Answer and work down below. Let me know if you have any questions

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What is the first derivative of r with respect to t (i.e., differentiate r with respect to t)? r = 5/(t2)Note: Use ^ to show exp
adelina 88 [10]

Answer:

The first derivative of r(t) = 5\cdot t^{-2} (r(t)=5*t^{-2}) with respect to t is r'(t) = -10\cdot t^{-3} (r'(t) = -10*t^{-3}).

Step-by-step explanation:

Let be r(t) = \frac{5}{t^{2}}, which can be rewritten as r(t) = 5\cdot t^{-2}. The rule of differentiation for a potential function multiplied by a constant is:

\frac{d}{dt}(c \cdot t^{n}) = n\cdot c \cdot t^{n-1}, \forall \,n\neq 0

Then,

r'(t) = (-2)\cdot 5\cdot t^{-3}

r'(t) = -10\cdot t^{-3} (r'(t) = -10*t^{-3})

The first derivative of r(t) = 5\cdot t^{-2} (r(t)=5*t^{-2}) with respect to t is r'(t) = -10\cdot t^{-3} (r'(t) = -10*t^{-3}).

5 0
3 years ago
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