3 years ago

もしもし！ Save me from this boredom. ( ˙▿˙ )

Mathematics

1 answer:

Anastasy [175]3 years ago

5 0

<em>uhh I have discord if you wanna talk kurxmii#6390</em>

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2 years ago

×>10 is a solution to which inequality a}×-7>3 b}×-7>3 c}x-7<3 d}x-7<3

NeX [460]

A) x - 7 > 3 |add 7 to both sides
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b) x - 7 > 3 |add 7 to both sides
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2 years ago

Let f be the function given by f(x)= (x-1)(x^2-4)/x^2-a. For what positive values of a is f continuous for all real numbers x?

9966 [12]

Answer:

a =1 and a=4.

Step-by-step explanation:

The function is

$f(x)=\frac{(x-1)(x^2-4)}{x^2-a}$

If we want f(x) to be continuous the denominator needs to be different to 0, otherwise f(x) will be indeterminate.

Now, for a a positive real we have that $x=\sqrt{a}$ will annulate the denominator, i.e

$(\sqrt{a})^2-a = a-a = 0$ . But, if a = 1 we have:

$f(x)=\frac{(x-1)(x^2-4)}{x^2-1} = \frac{(x-1)(x^2-4)}{(x-1)(x+1)}=\frac{(x^2-4)}{x+1}$

so, the value $x=\sqrt{a} = \sqrt{1} = 1$ won't annulate the denominator.

Now, for a = 4 we have:

$f(x)=\frac{(x-1)(x^2-4)}{x^2-4} = x-1$

so, the value $x=\sqrt{a} = \sqrt{4} = 2$ won't annulate the denominator.

In conclusion, for a=1 or a=1, the function will be continuos for all real numbers, since the denominator will never be 0.

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3 years ago

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Answer:

D. (3; -4)

Step-by-step explanation:

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2 years ago

the length of the two legs of a right triangle are 2 square root 3 and 5 what is the length of the hypotenuse

pashok25 [27]

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3 years ago