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noname [10]
3 years ago
10

Assuming x > 0, which of these expressions is equivalent to 11 times the square root of 245 x to the third plus 9 times the s

quare root of 45 x to the third? ?
5 x times the square root of 104 x

20 times the square root of 290 x to the sixth

20 x times the square root of 290 x

104 x times the square root of 5 x
Mathematics
2 answers:
boyakko [2]3 years ago
8 0
Writing the expression mathematically:
11√(245x³)  +  9√(45x³)
11√(7² * 5 * x² * x) + 9√(3² * 5 * x² * x)
77x√(5x) + 27x√(5x)
104x√(5x)

104x times the square root of 5x
The last option is correct.
ehidna [41]3 years ago
4 0
The answer is <span>104 x times the square root of 5 x 
</span>
<span>11 times the square root of 245 x to the third plus 9 times the square root of 45 x to the third is:
</span>11 \sqrt{245 x^{3} } +9 \sqrt{45 x^{3} } =11 \sqrt{245}  *\sqrt{ x^{3} } +9 \sqrt{45} *\sqrt{ x^{3} }= \\  \\ =11 \sqrt{5*49} * \sqrt{ x^{2} *x} +9 \sqrt{5*9} * \sqrt{ x^{2} *x}= \\  \\ =11* \sqrt{5}  * \sqrt{49} * \sqrt{ x^{2}} * \sqrt{x} +9* \sqrt{5} * \sqrt{9} * \sqrt{ x^{2} }* \sqrt{x} = \\  \\ &#10;=11* \sqrt{5} *7*x* \sqrt{x} +9* \sqrt{5} *3*x* \sqrt{x} = \\  \\ &#10;=77x* \sqrt{5*x} +27x* \sqrt{5*x}  = \\  \\ &#10;=77x \sqrt{5x} +27x \sqrt{5x} = \\  \\ &#10;=104x \sqrt{5x}
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11/ 42 Marks
amm1812

Answer:

(a)x=128 degrees

(b)\angle APD\neq 116^\circ

Step-by-step explanation:

\angle ABD+\angle CBD =180^\circ$ (Linear Postulate)\\\angle ABD+116^\circ =180^\circ\\\angle ABD=180^\circ-116^\circ\\\angle ABD=64^\circ

(a) Now:

\angle AOD=2 \times \angle ABD$ (Angle at the centre is twice the  angle at the circumference)\\x=2 \times 64^\circ\\x=128^\circ

(b)

\angle APD=2 \times \angle ABD$ (Inscribed Angle Theorem)\\\angle APD=2 \times 64^\circ\\ \angle APD=128^\circ

\angle APD\neq 116^\circ

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