2 years ago

In the definition of fractional numbers, b ≠ 0. Why?

Mathematics

1 answer:

Nataliya [291]2 years ago

6 0

I have no idea. Just commenting to get the answer when someone else knows it

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Mkey [24]

Are you familiar with that?

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amp = 4

and

period=2 pi /4 = pi /2

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2 years ago

Is the ratio 4.2:1.5 proportional to the ratio 12:5

dezoksy [38]

Yes this is proportional!

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3 years ago

HELP FAST PLEASE!!!<br> Show that sin(x+pi)=-sinx

I am Lyosha [343]

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2 years ago

Read 2 more answers

F() = x2 + 1<br> g(x) = 5 - x<br> (f+g)(x) =

frutty [35]

Answer:

(f+g)(x) = x² - x + 6

Step-by-step explanation:

We can find (f+g)(x) by adding f(x) and g(x).

f(x) = x² + 1

g(x) = 5 - x

(f+g)(x) = f(x) + g(x)

(f+g)(x) = (x² + 1) + (5 - x)

(f+g)(x) = x² + 1 + 5 - x

(f+g)(x) = x² - x + 6

4 0

2 years ago

Someone can help??

oee [108]

Answer:

Step-by-step explanation:

let cos^{-1}x=t

cos t=x

when x=1,cos t=1=cos 0

t \rightarrow 0

$\lim_{x \to 1} \frac{1-\sqrt{x}}{(cos ^{-1}x)^2 } \\= \lim_{t \to 0} \frac{1-\sqrt{cos~t}}{t^2} \times \frac{1+\sqrt{cos~t}}{1+\sqrt{cos ~t}} \\= \lim_{t \to 0} \frac{1-cos~t}{t^2(1+\sqrt{cos~t})}} \\= \lim_{t \to 0 }\frac{2 sin^2~\frac{t}{2}}{t^2(1+\sqrt{cos~t})}} \\= 2\lim_{t \to 0 }(\frac{sin~t/2}{\frac{t}{2} })^2 \times \frac{1}{4} \times \lim_{t \to 0 }\frac{1}{1+\sqrt{cos~t}} \\=\frac{2}4} \times 1^2 \times \frac{1}{1+\sqrt{cos~0}} \\=\frac{1}{2} \times \frac{1}{1+1} \\=\frac{1}{4}$

4 0

3 years ago