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3 years ago

What is the soloution of x^2-1/x^2+5x+4<_0

Mathematics

1 answer:

JulijaS [17]3 years ago

6 0

You have the following inequality given in the problem shown above:

(x^2-1)/(x^2+5x+4)≤0

1. To solve it, you must factor it and then you must make the study of the signs.

2. Once you do the proccedure mentioned above, you obtain the following solutions:

-4<x<-1

-1<x≤1

3. You can graph the inaquality given in the problem, as you can see in the figure attached.

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If the null space of a 4 x 9 matrix is 5-dimensional, what is the dimension of the column space of the matrix? Explain your answ

Naily [24]

Answer : Dimension of column A is also be 4 whereas the two vector basis lie in R⁴.

The smallest possible dimension of Nul A would be zero.

Step-by-step explanation:

Since we have given that

A is matrix of 4 x 9 .

so, Number of rows = 4

Number of columns = 9

Nul A = 5

It means that Rank of A would be 9 - 5 =4

So, rank A = 4

Thus, dimension of column A is also be 4 whereas the four vector basis lie in R⁴.

So, dim Col A = 4

If A is 7 x 3 matrix.

So, we know that

rank A + dim (null A) = 3

so, it is possible to have rank A = 3 so the dim col A should be 3

Then the smallest possible dimension of Nul A would be zero.

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3 years ago

A uranium mining town reported population declines of 3.2%, 5.2%, and 4.7% for the three successive five-year periods 1985–89, 1

KATRIN_1 [288]

Answer:

Step-by-step explanation:

Heres the complete question:

A uranium mining town reported population declines of 3.2%, 5.2%, and 4.7% for the three successive five-year periods 1985–89, 1990–94, and 1995–99. If the population at the end of 1999 was 9,320:

How many people lived in the town at the beginning of 1985? (Round your answer to the nearest whole number.)

solution:

Let the population of the town at the beginning of 1985 be P. Then, given that in the first five-year period the population declined by 3.2%, i.e., 0.032, the population of the town at the end of 1989 would be

(1 – 0.032)P = 0.968P.

Again, given that in the second five-year period the population declined by 5.2%, i.e., 0.052, the population of the town at the end of 1994 would be

(1 – 0.052)(0.968P) = 0.948 x 0.968P = 0.917664P.

Finally, given that in the third five-year period the population declined by 4.7%, i.e., 0.047, the population of the town at the end of 1999 would be

(1 – 0.047)(0.917664P) = 0.874533792P.

We are given, 0.874533792P = 9320 or

P = 9320/0.874533792 = 10657.11.

Thus, 10657 people lived in the town at the beginning of 1985

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4 years ago

Mark had 312 books. He has 11 times more fiction than nonfiction. How many fiction does he have

andrew11 [14]

He has about 28 fiction books. Since 312 divided by 11 is equal to 28.3636363636, you can't have a decimal number for the amount of books, so you round down to 28.

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3 years ago

Roberto tool a test that contained 25 questions. He received an 88% on the test. how many questions did he answer correctly?

iren [92.7K]

He got 22/25 questions right because

if you put X as the number of questions he got right out of 25 and equal it to 88 out of 100 you can cross multiply to get X

x/25=88/100
100x=25×88

25×88=2200
100x=2200

divide 100 on both sides
therefore
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How thick is the crust layer of the earth in kilometers?

Aleonysh [2.5K]

The earths crust depth in kilometers is 6.4 kilometers I hoped this helped

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3 years ago