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seropon [69]
3 years ago
5

When the fraction below is written as a repeating decimal, which digits are repeating?

Mathematics
1 answer:
Sergeeva-Olga [200]3 years ago
6 0

Answer:

A

Step-by-step explanation:

\frac{7}{11} ≈ 0.636363636...

with repeating digits 63

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So, when we multiply 20% by 5 (to get 100% = all the teachers), we get:
18 x 5 = 90 teachers in total
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Read 2 more answers
Find The derivative of:(cosx/1+sinx)^3​
forsale [732]

Answer:

-\frac{3 \cdot cos^2x}{(1+sinx)^3}

Step-by-step explanation:

y = (\frac{cosx}{1+sinx})^3\\\\\frac{dy}{dx} = 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{dy}{dx}(\frac{cosx}{1+sinx})                        [ y = x^n\  \  \ =>  \  \  \  \frac{dy}{dx} = b \cdot x^{n-1} \ ]

    = 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{(1+sin x(-sinx) - cosx(cosx)}{(1+sinx)^2}\\\\     [\ \frac{u}{v} = \frac{v \dcot u'- u \cdot v'}{v^2}\ ]

    = 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{-sin x-sin^2x- cos^2x}{(1+sinx)^2}\\\\= 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{-sin x- (sin^2x+ cos^2x)}{(1+sinx)^2}\\\\= 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{-sin x-1}{(1+sinx)^2}\\\\= 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{-1 \cdot(sin x+1)}{(1+sinx)^2}\\\\= 3 \cdot (\frac{cosx}{1+sinx})^2 \cdot \frac{-1}{(1+sinx)}\\\\

   = -3 \cdot \frac{cos^2x}{(1+sinx)^3}

   

5 0
3 years ago
Can someone answer this plzz
aev [14]

Answer:

16/2 = 8

Step-by-step explanation:

23+16= 39

39+16= 55

55+16= 71

1+2=3

3+2=5

5+2=7

3 0
3 years ago
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