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Damm [24]
1 year ago
15

Solve rational inequality x²+x-6/x²-3x-4≤0?​

Mathematics
2 answers:
Fittoniya [83]1 year ago
8 0

Answer:

Step-by-step explanation:

x

2

+

x

−

6

=

(

x

+

3

)

(

x

−

2

)

x

2

−

3

x

−

4

=

(

x

−

4

)

(

x

+

1

)

Each of the linear factors occurs precisely once, so the sign of the given rational expression will change at each of the points where one of the linear factors is zero. That is at:

x

=

−

3

,

−

1

,

2

,

4

Note that when

x

is large, the

x

2

terms will dominate the values of the numerator and denominator, making both positive.

Hence the sign of the value of the rational expression in each of the intervals

(

−

∞

,

−

3

)

,

(

−

3

,

−

1

)

,

(

−

1

,

2

)

,

(

2

,

4

)

and

(

4

,

∞

)

follows the pattern

+

−

+

−

+

. Hence the intervals

(

−

3

,

−

1

)

and

(

2

,

4

)

are both part of the solution set.

When

x

=

−

1

or

x

=

4

, the denominator is zero so the rational expression is undefined. Since the numerator is non-zero at those values, the function will have vertical asymptotes at those points (and not satisfy the inequality).

When

x

=

−

3

or

x

=

2

, the numerator is zero and the denominator is non-zero. So the function will be zero and satisfy the inequality at those points.

Hence the solution is:

x

∈

[

−

3

,

−

1

)

∪

[

2

,

4

)

graph{(x^2+x-6)/(x^2-3x-4) [-10, 10, -5, 5]}

Aleonysh [2.5K]1 year ago
7 0

Answer:

x ∈ [-3;-1) ∪ [2;4)

Step-by-step explanation:

\frac{x^{2}+x-6 }{x^{2}-3x-4 }\leq 0\\=> \frac{(x-2)(x+3)}{(x+1)(x-4)} \leq 0\\

we have this board:

x                          -3                    -1                      2                     4          

x - 2          -                        -                    -           0        +                     +

x+3            -         0           +                   +                      +                     +

x+1             -                      -            0      +                      +                     +

x-4            -                        -                     -                     -            0      +

\frac{(x-2)(x+3)}{(x+1)(x-4)}   +          0          -             ||        +         0        -            ||        +

from the board

=>    x ∈ [-3;-1) ∪ [2;4)

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the cross sectional area of a cylinder is 614 cm 2 the height is 8cm what is the volume of the cylinder​
Elina [12.6K]

Answer: Volume of the cylinder​ = 4912\ cm^2

Step-by-step explanation:

Volume of cylinder = Area of base x height

Also, area of cross section = Area of base

Volume of cylinder =  area of cross section x height

Given:  Area of cross section = 614\ cm^2

height = 8 cm

Volume of cylinder = 614 x 8

= 4912 \ cm^2

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6 0
1 year ago
The aquarium at Sea Critters Depot contains 140 fish. Eighty of these fish are green swordtails (44 female and 36 male) and 60 a
maxonik [38]

Given that:

Total number of fish = 140

Fish are green swordtails female = 44

Fish are green swordtails male = 36

Fish are orange swordtails female = 36

Fish are orange swordtails male = 24

Solution:

A. We have to find the probability that the selected fish is a green swordtail.

\text{P(green swordtail)}=\dfrac{\text{Total green swordtail fish}}{\text{Total fish}}

\text{P(green swordtail)}=\dfrac{80}{140}

\text{P(green swordtail)}=\dfrac{4}{7}

Therefore, the probability that the selected fish is a green swordtail is \dfrac{4}{7}.

B.  We have to find the probability that the selected fish is male.

\text{P(Male fish)}=\dfrac{\text{Total male fish}}{\text{Total fish}}

\text{P(Male fish)}=\dfrac{36+24}{140}

\text{P(Male fish)}=\dfrac{60}{140}

\text{P(Male fish)}=\dfrac{3}{7}

Therefore, the probability that the selected fish is a male, is \dfrac{3}{7}.

C. We have to find the probability that the selected fish is a male green swordtail.

\text{P(Male green swordtail)}=\dfrac{\text{Total male green swordtail fish}}{\text{Total fish}}

\text{P(Male green swordtail)}=\dfrac{36}{140}

\text{P(Male green swordtail)}=\dfrac{9}{35}

Therefore, probability that the selected fish is a male green swordtail is \dfrac{9}{35}.

D.

We have to find the probability that the selected fish is either a male or a green swordtail.

\text{P(Male or green swordtail)}=\dfrac{\text{Total male or green swordtail fish}}{\text{Total fish}}

\text{P(Male or green swordtail)}=\dfrac{44+36+24}{140}

\text{P(Male or green swordtail)}=\dfrac{96}{140}

\text{P(Male or green swordtail)}=\dfrac{24}{35}

Therefore, the probability the selected fish is either a male or a green swordtail is \dfrac{24}{35}.

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Step-by-step explanation:

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