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iren2701 [21]
3 years ago
11

PLEASE HELP (algebra 1)

Mathematics
1 answer:
padilas [110]3 years ago
6 0
A) 6(4)^1-6(4)^0 = 24-6 = 18
    6(4)^3-6(4)^2 = 384-96 = 288
b) It is 27.11 times difference. This is because g(x) is an exponential function and so as x increases, g(x) increases faster.
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Find the derivative of sinx/1+cosx, using quotient rule​
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Answer:

f'(x) = -1/(1 - Cos(x))

Step-by-step explanation:

The quotient rule for derivation is:

For f(x) = h(x)/k(x)

f'(x) = \frac{h'(x)*k(x) - k'(x)*h(x)}{k^2(x)}

In this case, the function is:

f(x) = Sin(x)/(1 + Cos(x))

Then we have:

h(x) = Sin(x)

h'(x) = Cos(x)

And for the denominator:

k(x) = 1 - Cos(x)

k'(x) = -( -Sin(x)) = Sin(x)

Replacing these in the rule, we get:

f'(x) = \frac{Cos(x)*(1 - Cos(x)) - Sin(x)*Sin(x)}{(1 - Cos(x))^2}

Now we can simplify that:

f'(x) = \frac{Cos(x)*(1 - Cos(x)) - Sin(x)*Sin(x)}{(1 - Cos(x))^2} = \frac{Cos(x) - Cos^2(x) - Sin^2(x)}{(1 - Cos(x))^2}

And we know that:

cos^2(x) + sin^2(x) = 1

then:

f'(x) = \frac{Cos(x)- 1}{(1 - Cos(x))^2} = - \frac{(1 - Cos(x))}{(1 - Cos(x))^2} = \frac{-1}{1 - Cos(x)}

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3 years ago
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A rhombus is related to a “Parallelogram”

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hope this helped <3
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