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babunello [35]
3 years ago
6

42r^2s^4/14r^2s^2z^4

Mathematics
2 answers:
ser-zykov [4K]3 years ago
8 0
Answer 3s^2 / z^4

Subtract the exponents four.s on top cancel with two on bottom leaves two s on top
shutvik [7]3 years ago
4 0
12 is your answer 
you are welcoe
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The first step in the process for factoring the trinomial x^2-3x-40 is to
jenyasd209 [6]

(x-8)(x+5)

First thing you do is put the X's into the parentheticals because it is x^2 meaning there are 2 factors with x's. Then, you wonder what factors of -40 could add to get -3. In this case, they are -8 and 5. -8 times 5 is -40 and -8 +5 is -3. Then, you just add -8 and 5 beside the x's in the parentheticals.

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The derivative of y with respect to x, when y = 3e^5x - 2 is:
Ede4ka [16]
Answer: 15e^5x

Step - by - step

y=3e^5x - 2

By the sum rule, the derivative of 3e^5x - 2 with respect to x is d/dx [ 3e^5x ] + d/dx [-2].

d/dx [ 3e^5x ] + d/dx [ -2 ]

Evalute d/dx [ 3e^5x ]

Since 3 is constant with respect to x , the derivative of 3e^5x with respect to x is
3 d/dx [ e^5x ].

3 d/dx [ e^5x ] + d/dx [ -2 ]

Differentiate using the chain rule, which states that d/dx [ f(g(x))] is f' (g(x)) g' (x) where f(x) = e^x and g(x) = 5x.

To apply the Chain Rule, set u as 5x.
3 ( d/du [ e^u] d/dx [5x] ) + d/dx [ -2]

Differentiate using the Exponential rule which states that d/du [ a^u ] is a^u ln(a) where a=e.

3( e^u d/dx[5x] ) + d/dx [ -2 ]
Replace
3(e^5x d/dx [5x] ) + d/dx [ -2 ]
3(e^5x( 5 d/dx [x] )) + d/dx [ -2 ]

Diffentiate using the Power Rule which states that d/dx [x^n] is nx^n-1 where n=1.
3(e^5x(5*1)) + d/dx [-2]
3 ( e^5x * 5 ) + d/dx [-2]

Multiply 5 by 3
15e^5x + d/dx [-2]

Since -2 is constant with respect to x, the derivative of -2 with respect to x is 0.
15e^5x + 0
15e^5x
5 0
10 months ago
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