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Papessa [141]
3 years ago
7

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

Mathematics
1 answer:
LekaFEV [45]3 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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hope it helps

8 0
3 years ago
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The graph below represents the solution set of which inequality?
natulia [17]

Answer:

option: B (x^2+2x-8) is correct.

Step-by-step explanation:

We are given the solution set as seen from the graph as:

(-4,2)

1)

On solving the first inequality we have:

x^2-2x-8

On using the method of splitting the middle term we have:

x^2-4x+2x-8

⇒  x(x-4)+2(x-4)=0

⇒ (x+2)(x-4)

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

x+2>0 and x-4

i.e. x>-2 and x<4

so we have the region as:

(-2,4)

Case 2:

x+2 and x-4>0

i.e. x<-2 and x>4

Hence, we did not get a common region.

Hence from both the cases we did not get the required region.

Hence, option 1 is incorrect.

2)

We are given the second inequality as:

x^2+2x-8

On using the method of splitting the middle term we have:

x^2+4x-2x-8

⇒ x(x+4)-2(x+4)

⇒ (x-2)(x+4)

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

x-2>0 and x+4

i.e. x>2 and x<-4

Hence, we do not get a common region.

Case 2:

x-2 and x+4>0

i.e. x<2 and x>-4

Hence the common region is (-4,2) which is same as the given option.

Hence, option B is correct.

3)

x^2-2x-8>0

On using the method of splitting the middle term we have:

x^2-4x+2x-8>0

⇒ x(x-4)+2(x-4)>0

⇒ (x-4)(x+2)>0

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

x+2>0 and x-4>0

i.e. x>-2 and x>4

Hence, the common region is (4,∞)

Case 2:

x+2 and x-4

i.e. x<-2 and x<4

Hence, the common region is: (-∞,-2)

Hence, from both the cases we did not get the desired answer.

Hence, option C is incorrect.

4)

x^2+2x-8>0

On using the method of splitting the middle term we have:

x^2+4x-2x-8>0

⇒ x(x+4)-2(x+4)>0

⇒ (x-2)(X+4)>0

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

x-2 and x+4

i.e. x<2 and x<-4

Hence, the common region is: (-∞,-4)

Case 2:

x-2>0 and x+4>0

i.e. x>2 and x>-4.

Hence, the common region is: (2,∞)

Hence from both the case we do not have the desired region.

Hence, option D is incorrect.




5 0
3 years ago
The March utility bills (in dollars) of 30 homeowners are listed below. 44 38 41 50 36 36 43 42 49 48 35 40 37 41 43 50 45 45 39
Ulleksa [173]

Step-by-step explanation:

What we will do is create a table with the data they give us, the first thing will be to organize it from smallest to largest, like this:

33, 35, 35, 36, 36, 36, 37, 38, 38, 39, 40, 40, 41, 41, 41, 42, 42, 43, 43, 43, 44, 45, 45, 47, 48, 48, 49, 50, 50, 50

Now we know that the smallest number is 33 and the largest is 50, to create 5 ranges, we will calculate the difference and divide by 5.

(50 - 33) / 5 = 3.4

Therefore we will make ranges of 3 and 4, values, like this

Rank 1: 33 - 36

Rank 2: 37 - 39

Rank 3: 40 - 43

Rank 4: 44 - 46

Rank 5: 47 - 50

We will calculate the frequency distribution of values in each range:

Rank 1: 6

Rank 2: 4

Rank 3: 10

Rank 4: 3

Rank 5: 7

5 0
3 years ago
Bradley cut a square hole out of a block of wood in wood shop. if the block was cube-shaped with side lengths of 8 inches, and t
borishaifa [10]
8^3-3^3
512-27
485 cubic inches
3 0
4 years ago
BRAINLIEST!!<br> 17. Complete the statement about parallelogram ABCD.<br> ∠BDC ____
Vsevolod [243]

Answer:

∠ABD; Alternate Interior angles are congruent

Step-by-step explanation:

Sides AB and DC are congruent and parallel.

The transversal is line BD.

So considering this things, ∠BDC and ∠ABD are congruent through alternate interior angle rule

6 0
3 years ago
Read 2 more answers
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