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bezimeni [28]
2 years ago
12

39-42 Differentiate

la"> and find the domain of f\text{.}
40. f(x)=\ln \left(x^{2}-2 x\right)
Mathematics
1 answer:
MAVERICK [17]2 years ago
4 0

\text{Given that,}\\\\f(x) = \ln (x^2 - 2x)\\\\f'(x) = \dfrac  1{x^2 -2x} \cdot (2x - 2) = \dfrac{2(x-1)}{x(x-2)}\\\\\\\text{domain f(x): }   \{x| x < 0, x > 2 \}

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Naya [18.7K]

Answer:

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3 years ago
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MariettaO [177]

Answer:

5.14

Step-by-step explanation:

first you divide 7 by 4 then you get 1.75, then you want to divide 1.75 by 1.75 to get it to 0,  then you divide 9 by 1.75 so b=5.14

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3 years ago
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IrinaK [193]

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6 0
3 years ago
Find sin^4(a) + cos^4(a), if cos(a) + sin(a) = 1/3
Nadya [2.5K]

Answer:

49/81

Step-by-step explanation:

[cos(a) + sin(a)]^2 = (1/3)^2

(cos(a))^2 + 2sin(a)cos(a) + (sin(a))^2 = 1/9

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

1 + 2sin(a)cos(a) = 1/9

2sin(a)cos(a) = -8/9

sin(a)cos(a) = -4/9

[cos(a) + sin(a)]^4 = (1/3)^4 = 1/81

(cos(a))^4 + 4sin(a)×(cos(a))^3 + 6×(sin(a))^2×(cos(a))^2 + 4(sin(a))^3×cos(a) + (sin(a))^4 = 1/81

(cos(a))^4 + (sin(a))^4 + 4sin(a)cos(a)((cos(a))^2 + (sin(a))^2) + 6(sin(a)cos(a))^2 = 1/81

cos(a))^4 + (sin(a))^4 + 4sin(a)cos(a)(1) + 6(sin(a)cos(a))^2 = 1/81

(cos(a))^4 + (sin(a))^4 + 4(-4/9) +6((-4/9)^2) = 1/81

(cos(a))^4 + (sin(a))^4 - 16/9 + 6(16/81) = 1/81

(cos(a))^4 + (sin(a))^4 = 1/81 + 16/9 - 6(16/81)

(cos(a))^4 + (sin(a))^4 = 49/81

3 0
3 years ago
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