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Elis [28]
3 years ago
12

Divide 59ml into two parts so that one part is 7ml less than five times the other parts​

Mathematics
1 answer:
VashaNatasha [74]3 years ago
7 0

Answer:  The two parts are 48 ml and 11 ml.

Step-by-step explanation:

Let x= The quantity in the other part ( in ml).

Then, the quantity in the first part = 5x-7

Since , Whole part = Sum of both parts

According to the question , we have

59=5x-7+x\\\\

Add 7 on both sides , we get

66=6x

Divide both sides by 6 , we get

11=x

i.e. Value of  x= 11

Therefore , the quantity in the first part = 5(11)-7=48 ml

The quantity in the other part = 11 ml.

Hence, the two parts are 48 ml and 11 ml.

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Simplify completely <br> (X^2+x-12/x^2-x-20)/(3x^2–24x+45/12x^2-48-60)
Rudik [331]

Answer:

(4 (x^4 - 20 x^2 - 12))/(3 x^2 (9 x^2 - 32 x - 144))

Step-by-step explanation:

Simplify the following:

(x^2 + x - x - 20 - 12/x^2)/((15 x^2)/4 + 3 x^2 - 24 x - 60 - 48)

Hint: | Put the fractions in x^2 + x - x - 20 - 12/x^2 over a common denominator.

Put each term in x^2 + x - x - 20 - 12/x^2 over the common denominator x^2: x^2 + x - x - 20 - 12/x^2 = x^4/x^2 + x^3/x^2 - x^3/x^2 - (20 x^2)/x^2 - 12/x^2:

(x^4/x^2 + x^3/x^2 - x^3/x^2 - (20 x^2)/x^2 - 12/x^2)/((45 x^2)/12 + 3 x^2 - 24 x - 60 - 48)

Hint: | Combine x^4/x^2 + x^3/x^2 - x^3/x^2 - (20 x^2)/x^2 - 12/x^2 into a single fraction.

x^4/x^2 + x^3/x^2 - x^3/x^2 - (20 x^2)/x^2 - 12/x^2 = (x^4 + x^3 - x^3 - 20 x^2 - 12)/x^2:

((x^4 + x^3 - x^3 - 20 x^2 - 12)/x^2)/((45 x^2)/12 + 3 x^2 - 24 x - 60 - 48)

Hint: | Group like terms in x^4 + x^3 - x^3 - 20 x^2 - 12.

Grouping like terms, x^4 + x^3 - x^3 - 20 x^2 - 12 = x^4 - 20 x^2 - 12 + (x^3 - x^3):

(x^4 - 20 x^2 - 12 + (x^3 - x^3))/(x^2 ((45 x^2)/12 + 3 x^2 - 24 x - 60 - 48))

Hint: | Look for the difference of two identical terms.

x^3 - x^3 = 0:

((x^4 - 20 x^2 - 12)/x^2)/((45 x^2)/12 + 3 x^2 - 24 x - 60 - 48)

Hint: | In (45 x^2)/12, the numbers 45 in the numerator and 12 in the denominator have gcd greater than one.

The gcd of 45 and 12 is 3, so (45 x^2)/12 = ((3×15) x^2)/(3×4) = 3/3×(15 x^2)/4 = (15 x^2)/4:

(x^4 - 20 x^2 - 12)/(x^2 (15 x^2/4 + 3 x^2 - 24 x - 60 - 48))

Hint: | Put the fractions in (15 x^2)/4 + 3 x^2 - 24 x - 60 - 48 over a common denominator.

Put each term in (15 x^2)/4 + 3 x^2 - 24 x - 60 - 48 over the common denominator 4: (15 x^2)/4 + 3 x^2 - 24 x - 60 - 48 = (15 x^2)/4 + (12 x^2)/4 - (96 x)/4 - 240/4 - 192/4:

(x^4 - 20 x^2 - 12)/(x^2 (15 x^2)/4 + (12 x^2)/4 - (96 x)/4 - 240/4 - 192/4)

Hint: | Combine (15 x^2)/4 + (12 x^2)/4 - (96 x)/4 - 240/4 - 192/4 into a single fraction.

(15 x^2)/4 + (12 x^2)/4 - (96 x)/4 - 240/4 - 192/4 = (15 x^2 + 12 x^2 - 96 x - 240 - 192)/4:

(x^4 - 20 x^2 - 12)/(x^2 (15 x^2 + 12 x^2 - 96 x - 240 - 192)/4)

Hint: | Group like terms in 15 x^2 + 12 x^2 - 96 x - 240 - 192.

Grouping like terms, 15 x^2 + 12 x^2 - 96 x - 240 - 192 = (12 x^2 + 15 x^2) - 96 x + (-192 - 240):

(x^4 - 20 x^2 - 12)/(x^2 ((12 x^2 + 15 x^2) - 96 x + (-192 - 240))/4)

Hint: | Add like terms in 12 x^2 + 15 x^2.

12 x^2 + 15 x^2 = 27 x^2:

(x^4 - 20 x^2 - 12)/(x^2 (27 x^2 - 96 x + (-192 - 240))/4)

Hint: | Evaluate -192 - 240.

-192 - 240 = -432:

(x^4 - 20 x^2 - 12)/(x^2 (27 x^2 - 96 x + -432)/4)

Hint: | Factor out the greatest common divisor of the coefficients of 27 x^2 - 96 x - 432.

Factor 3 out of 27 x^2 - 96 x - 432:

(x^4 - 20 x^2 - 12)/(x^2 (3 (9 x^2 - 32 x - 144))/4)

Hint: | Write ((x^4 - 20 x^2 - 12)/x^2)/((3 (9 x^2 - 32 x - 144))/4) as a single fraction.

Multiply the numerator by the reciprocal of the denominator, ((x^4 - 20 x^2 - 12)/x^2)/((3 (9 x^2 - 32 x - 144))/4) = (x^4 - 20 x^2 - 12)/x^2×4/(3 (9 x^2 - 32 x - 144)):

Answer: (4 (x^4 - 20 x^2 - 12))/(3 x^2 (9 x^2 - 32 x - 144))

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