Help on number 3 only use Pythagorean thym I give brainliest
1 answer:
Answer: 10.8
Step-by-step explanation: the pythagorean theorem is A sqaured or A^2 + B^2 or B squared equals C^2 or C squared so 10 squared as is the first number is 100 and 4 squared as is the second number is 16 add the two to get 116 but since your answer is squared you need to take the square root which means your answer is 10.7703296143 but since your rounding to the tenths place your answer is 10.8
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Answer:
Solution is x= 0 , x = \frac{1}{14} , x = \frac{-1}{12}
Step-by-step explanation:
Given, equation is \sqrt[3]{15x-1} + \sqrt[3]{13x+1} = 4\sqrt[3]{x} →→→ (1)
Now, by cubing the equation on both sides, we get
( \sqrt[3]{15x-1} + \sqrt[3]{13x+1} )³ = (4\sqrt[3]{x})³
⇒ (15x-1) + (13x+1) + 3×\sqrt[3]{15x-1}× \sqrt[3]{13x+1} (\sqrt[3]{15x-1} + \sqrt[3]{13x+1}) = 64 x.
(since (a+b)³ = a³ + b³ + 3ab(a+b) ).
⇒ 28x + 3×\sqrt[3]{15x-1}× \sqrt[3]{13x+1} (\sqrt[3]{15x-1} + \sqrt[3]{13x+1}) = 64x.
(since from (1), \sqrt[3]{15x-1} + \sqrt[3]{13x+1} = 4\sqrt[3]{x} )
⇒ 12×\sqrt[3]{15x-1}× \sqrt[3]{13x+1} (\sqrt[3]{15x-1}×\sqrt[3]{x}= 36x.
⇒ 3x = .
Now, once again cubing on both sides, we get
(3x)³ = ()³.
⇒ 27x³ = (15x-1)(13x+1)(x).
⇒ 27x³ = 195x³ + 2x² - x
⇒ 168x³ + 2x² - x = 0
⇒ x(168x² + 2x -1) = 0
⇒ by, solving the equation we get ,
x = 0 ; x = \frac{1}{14} ; x = \frac{-1}{12}
therefore, solution is x= 0 , x = \frac{1}{14} , x = \frac{-1}{12}