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muminat
4 years ago
10

I need help with this problem

Mathematics
2 answers:
GaryK [48]4 years ago
6 0
The answer would be 7.28 p.m.

Hope It helps

21+42+7=63

8.31-0.63-0.4=7.28

morpeh [17]4 years ago
4 0
42+14+7=63 min =1 hour 3 minutes. Subtract that from 8:31 and you get 7:28 pm
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Step-by-step explanation:

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3 years ago
How can you tell if 78 is divisible by 2 3 OR 6
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Step-by-step explanation:

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8 0
3 years ago
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FinnZ [79.3K]

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Step-by-step explanation:

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7 0
3 years ago
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D:x^2-4\not=0\\&#10;D:x^2\not=4\\&#10;D:x\not=-2 \wedge x\not =2\\\\&#10;\displaystyle&#10;\lim_{x\to-2^-}\dfrac{(x-9)(x+7)}{x^2-4}=\\&#10;\dfrac{(-2-9)(-2+7)}{(-2^-)^2-4}=\dfrac{-11\cdot5}{4^+-4}=\dfrac{-55}{0^+}=-55\cdot\infty=-\infty\\&#10;\dfrac{(-2-9)(-2+7)}{(-2^+)^2-4}=\dfrac{-11\cdot5}{4^--4}=\dfrac{-55}{0^-}=-55\cdot(-\infty)=\infty&#10;

\displaystyle&#10;\lim_{x\to2^-}\dfrac{(x-9)(x+7)}{x^2-4}=\\&#10;\dfrac{(2-9)(2+7)}{(2^-)^2-4}=\dfrac{-7\cdot9}{4^--4}=\dfrac{-63}{0^-}=-63\cdot(-\infty)=\infty\\&#10;\dfrac{(2-9)(2+7)}{(2^+)^2-4}=\dfrac{-7\cdot9}{4^+-4}=\dfrac{-63}{0^+}=-63\cdot\infty=-\infty\\

So, the vertical asymptotes are x=\pm 2
5 0
4 years ago
Read 2 more answers
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