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2 years ago

What is the domain of the function y=%/x-1? O- o -1 < x < oo 0 O 1

Mathematics

1 answer:

LekaFEV [45]2 years ago

6 0

Answer:

$-\infty < x < \infty$

Step-by-step explanation:

Given

$y = \sqrt[3]{x - 1}$

Required

The domain

The given function is cubic root; there are no restrictions on cubic root functions

Hence, the domain is:

$-\infty < x < \infty$

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What is the coefficient of n^2 in the sum of -5^4+4n^2-n+5 and 7n^4-5n^3-9n^2+3n+2

pickupchik [31]

Hello :
-5^4+4n^2-n+5 : the coefficient of n^2 is : 4
7n^4-5n^3-9n^2+3n+2 : he coefficient of n^2 is :-9

4 0

2 years ago

Thanks for the help!

Bogdan [553]

Hope you could understand.

If you have any query, feel free to ask.

5 0

2 years ago

PLEASE HELP QUICK!! WILL OFFER 100 POINTS TO THE FIRST ONE THAT ANSWERS

Reil [10]

Given

$x+1 = \sqrt{7x+15}$

We have to set the restraint

$x+1\geq 0 \iff x \geq -1$

because a square root is non-negative, and thus it can't equal a negative number. With this in mind, we can square both sides:

$(x+1)^2=7x+15 \iff x^2+2x+1=7x+15 \iff x^2-5x-14 = 0$

The solutions to this equation are 7 and -2. Recalling that we can only accept solutions greater than or equal to -1, 7 is a feasible solution, while -2 is extraneous.

Similarly, we have

$x-3 = \sqrt{x-1}+4 \iff x-7=\sqrt{x-1}$

So, we have to impose

$x-7\geq 0 \iff x \geq 7$

Squaring both sides, we have

$(x-7)^2=x-1 \iff x^2-14x+49=x-1 \iff x^2-15x+50 = 0$

The solutions to this equation are 5 and 10. Since we only accept solutions greater than or equal to 7, 10 is a feasible solution, while 5 is extraneous.

7 0

2 years ago

Write in simplest form. n+1/7=2/3

ziro4ka [17]

I hope this helps you

n=2/3-1/7

n=2.7/3.7-1.3/7.3

n=14/21-3/21

n=11/21

7 0

2 years ago

F+0.2=−3 what does f equal

Sav [38]

f + 0.2 =-3

f = -3 -0.2

f = -3.2