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Elodia [21]
2 years ago
11

What is 10 51/77 as a improper fraction

Mathematics
1 answer:
Natali5045456 [20]2 years ago
6 0

Answer: 821/77

Step-by-step explanation:

First multiply 10 by 77 (770) , then add 51 (821)

Set 821 as the numerator or top

Keep 77 as the denominator

Final answer is 821/77

Hope this helps!

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The numbers 55 and 55.
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During a birthday party, 1/4 drank fanta orange, 1/3 took coca cola 1/5 of the remainder drank sprite and the rest took krest wh
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1/12

Step-by-step explanation:

Since 1/4 drank fanta orange and 1/3 took coca cola, the. remainder will be:

= 1 - (1/4 + 1/3)

= 1 - 7/12

= 5/12

Therefore, 1/5 of the remainder drank sprite, the fraction that drank sprite will be:

= 1/5 × 5/12

= 1/12

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How much of the circle is shaded? Write your answer as a fraction in simplest form.
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11 months ago
Determine the coefficient of x^3 in the expansion of (1 – x)^5(1 + 1/x)^5
Mkey [24]

Notice that

(1 - <em>x</em>)⁵ (1 + 1/<em>x</em>)⁵ = ((1 - <em>x</em>) (1 + 1/<em>x</em>))⁵ = (1 - <em>x</em> + 1/<em>x</em> - 1)⁵ = (1/<em>x</em> - <em>x</em>)⁵

Recall the binomial theorem:

\displaystyle(a+b)^n = \sum_{k=0}^n\binom nk a^{n-k}b^k

Let <em>a</em> = 1/<em>x</em>, <em>b</em> = -<em>x</em>, and <em>n</em> = 5. Then

\displaystyle\left(\frac1x-x\right)^5 = \sum_{k=0}^5\binom5k\left(\frac1x\right)^{5-k}(-x)^k = \sum_{k=0}^5 (-1)^k\binom5k x^{2k-5}

We get an <em>x</em> ³ term for

2<em>k</em> - 5 = 3   ==>   2<em>k</em> = 8   ==>   <em>k</em> = 4

so that the coefficient would be

(-1)^4\dbinom54 = \boxed{5}

4 0
2 years ago
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