The quotient of 49 and d increased by the product of k and f?

Mathematics

1 answer:

stich3 [128]3 years ago

6 0

"d increased by the product of k and f" is d+kf

49/(d+kf)

The final answer is 49/(d+kf)~

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Hello!

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Solution:

The rationalisation factor for $\frac{1}{ a - \sqrt{b} }$ is $a + \sqrt{b}$

So, let us apply it here.

$\frac{1}{5 - \sqrt{2} }$

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Therefore, multiplying and dividing by 5 + √2, we have

$= \frac{1}{5 - \sqrt{2} } \times \frac{5 + \sqrt{2} }{5 + \sqrt{2} } \\ = \frac{5 + \sqrt{2} }{(5 - \sqrt{2})(5 + \sqrt{2} ) } \\ = \frac{5 + \sqrt{2} }{ {(5)}^{2} - ( \sqrt{2})^{2} } \\ = \frac{5 + \sqrt{2} }{25 - 2} \\ = \frac{5 + \sqrt{2} }{23}$

Answer:

$\frac{5 + \sqrt{2} }{23}$

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Minchanka [31]

Answer:

Step-by-step explanation:

Remember the last question? to find circumference you have to multiply the diameter by 3.14? well same here except you divide the circumference by 3.14 to get the diameter. so divide 18.84 ÷ 3.14 = 6

~R3V0

$d= \frac{C}{\pi }$