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3 years ago

If n is the term, what is the first integer value of n where the sequence 2n² is greater than 50?

Mathematics

1 answer:

Nezavi [6.7K]3 years ago

3 0

Answer:

The 6th term is first integer value of n greater then 50 in the 2n^2 sequence.

Step-by-step explanation:

2n^2 sequence:

2,8,18,32,50,72

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Natalija [7]

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Step-by-step explanation:

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2 years ago

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In triangle opq right angled at p op=7cm,oq-pq=1 determine the values of sinq and cosq

soldi70 [24.7K]

Answer:

see explanation

Step-by-step explanation:

let pq = x

given oq - pq = 1 then oq = 1 + x

Using Pythagoras' identity, then

(oq)² = 7² + x²

(1 + x)² = 49 + x² ( expand left side )

1 + 2x + x² = 49 + x² ( subtract 1 from both sides )

2x + x² = 48 + x² ( subtract x² from both sides )

2x = 48 ( divide both sides by 2 )

x = 24 ⇒ pq = 24

and oq = 1 + x = 1 + 24 = 25 ← hypotenuse

sinq = $\frac{opposite}{hypotenuse}$ = $\frac{7}{25}$

cosq = $\frac{adjacent}{hypotenuse}$ = $\frac{24}{25}$

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3 years ago

Tju [1.3M]

Answer:

Step-by-step explanation:

Rationalize the denominator of b. So, multiply the numerator and denominator by $\sqrt{x}$

$b = \frac{(1-2\sqrt{x}) *\sqrt{x}}{\sqrt{x}*\sqrt{x} }=\frac{1*\sqrt{x} -2\sqrt{x} *\sqrt{x} }{\sqrt{x} *\sqrt{x} }\\\\=\frac{\sqrt{x} -2x}{x}\\$

Now, find a +b

$a +b = \frac{2x+\sqrt{x} }{x}+\frac{\sqrt{x} -2x}{x}\\\\=\frac{2x+\sqrt{x} +\sqrt{x} -2x}{x}$

Combine like terms

$= \frac{2x-2x+\sqrt{x} +\sqrt{x} }{x}\\\\=\frac{2\sqrt{x} }{x}$

Now find (a + b)²

(a +b)² = $(\frac{2\sqrt{x} }{x})^{2}$

$= \frac{2^{2}*(\sqrt{x} )^{2}}{x^{2}}\\\\= \frac{4* x}{x^{2}}\\\\= \frac{4}{x}$

Hint: $\sqrt{x} *\sqrt{x} =\sqrt{x*x}=x$

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mars1129 [50]

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