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Ksenya-84 [330]
3 years ago
12

Convert these improper fractions into mixed numbers. 5/4

Mathematics
2 answers:
KiRa [710]3 years ago
8 0

Answer: 1  1/4

Step-by-step explanation: first we have to find the whole number then we divide the numerator and the denominator. then we are only interested in the whole numbers so we ignore the one to the right of the decimal. then we will used the whole number we calculated in step one and multiply it by the original denominator then is subtracted from original numerator. our next step is to simplify.  then when that down we need to calculate the greatest common factor.

tekilochka [14]3 years ago
4 0

Answer:

1 1/4

Step-by-step explanation:

4/4 + 1/4 = 5/4

4/4 = 1

so 1 1/4

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3 years ago
In Corpus Christi, Texas, 4 dolphins ate 28 pounds of fish in one day. At this rate how many pounds of fish could 18 dolphins ea
Anuta_ua [19.1K]

Answer:126 pounds

Step-by-step explanation:

The ratio from dolphins to pound of fish in one day is 1:7

So you have to multiply the ratio on both sides by 18

So you will get 18:126

7 0
3 years ago
Find the interval in which the function is positive x^2-x-6=y
Semmy [17]

Answer:

(-INFINITY , -2) (3 , INFINITY)

Function is positive when above the x axis

Function is below the x axis between -2 and 3

6 0
3 years ago
Read 2 more answers
How to solve a=9y+3yx
Katena32 [7]
A = 9y+3xy
take 3y as a common factor:
a = 3y(3+x)
this is the simplest form of this expression.
5 0
3 years ago
Need the answers plz
Shkiper50 [21]
1)
           x^2 + 4x - 5
y =   --------------------
              3x^2 - 12

Vertcial asym => find the x-values for which the equation is not defined and check whether the limit goes to + or - infinite

=> 3x^2 - 12 = 0 => 3x^2 = 12

=> x^2 = 12 / 3 = 4

=> x = +/-2

Limit of y when x -> 2(+) = ( 2^2 + 4(2) - 5) / 0 = - 1 / 0(+) =  ∞

Limit of y when x -> 2(-) = - 1 / (0(-) = - ∞

Limit of y when x -> - 2(+) = +∞

Limit of y when x -> - 2(-) = -

=> x = 2 and x = - 2 are a vertical asymptotes

Horizontal asymptote => find whether y tends to a constant value when x -> infinite of negative infinite

Limi of y when x -> - ∞ = 1/3

Lim of y when x -> +∞ = 1/3

=> Horizontal asymptote y = 1/3

x - intercept => y = 0

=>
          x^2 + 4x - 5
0 =   ------------------ => x ^2 + 4x  - 5 = 0
              3x^2 - 12

Factor x^2 + 4x - 5 => (x + 5) (x - 1) = 0 => x = - 5 and x = 1

=> x-intercepts x = - 5 and x = 1

Domain: all the real values except x = 2 and x = - 2

2)  y = - 2 / (x - 4) - 1

using the same criteria you get:

Vertical asymptote: x = 4

Horizontal asymptote: y = - 1

Domain:all the real values except x = 4

Range: all the real values except y = - 1

3) 

x/ (x + 2) + 7 / (x - 5) = 14 / (x^2 - 3x - 10)

factor x^2 - 3x - 10 => (x - 5)(x + 2)

Multiply both sides by (x - 5) (x + 2)

=> x(x - 5) + 7( x + 2) = 14

=> x^2 - 5x + 7x + 14 = 14

=> x^2 + 2x = 0

=> x(x + 2) = 0 => x = 0 and x = - 2 but the function is not defined for x = - 2 so it is not a solution => x = 0

Answer: x = 0
4 0
3 years ago
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