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sp2606 [1]
3 years ago
9

Please help me the denominator only

Mathematics
2 answers:
vazorg [7]3 years ago
7 0

Answer:

  1. 6/5
  2. 1/3
  3. 3/10
  4. 13/5
  5. 49/20
  6. 1/15

Step-by-step explanation:

You cannot subtract JUST a denominator. The subtraction sign is lower down because of the odd printing just like the equal signs.

1. The first thing you have to do is change this from a mixed fraction into an improper fraction in order to subtract. This becomes:

  • 12/5 -6/5 (This is done by multiplying the large number by the denominator and adding it to the exsisiting numerator)
  • 12-6=6 therefore  the answer is 6/5

2. This one requires you to multiply your denominator so that it is the same for both fractions. The least common denominator (LCD) for these fractions is 12

  • 7/12-1/4(3) (Because 3x4 is 12 which is the LCD)
  • 7/12-3/12 = 4/12 (this can be simplified)
  • 1/3

3. This is fairly straight-forward the denominators already match so just subtract the numerators

  • 6/10-3/10= 3/10

4. Because we have a full number we must make it into a fraction in order to subtract it and also turn the mixed fraction into an improper fraction

  • We'll start with the mixed because we've done that before multiplying 4 by 5. Add to that 20 the 2 from the original fraction to get 22/ 5
  • For the full number use the denominator of the number to determine the fraction. Multiply the full number by the denominator (5) to get 35 and place a 5 underneath and call it a day.
  • Therefore: 35/5-22/5=13/5

5. This problem requires multiple steps the first would be to make improper fractions then get a LCD and then of course subtract.

  • (6x8)+2=48+2=50 --> 50/8
  • (5x3)+4=15+4=19 --> 19/5
  • LCD 5:5,15,20,25,30,40 & LCD 8: 8,16,24,32,40 LCD=40
  • (50/8)x5-(19/5)x8 = 250/40-152/40 = 98/40 This can be simplified
  • 49/20

6. For this we will use another LCD.

  • LCD 3: 3,6,9,12,15 LCD 5: 5,10,15 LCD=15
  • (2/3)x5-(3/5)x3 = 10/15-9/15 = 1/15
viktelen [127]3 years ago
5 0

Answer:

1. 1/1/5

2. 1/3

3. 3/10

4. 2/3/5

5. 2/9/20

6. 1/15

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Solve the equation 2x^3-5x^2+x+2=0 given that 2 is a zero of f(x)= 2x^3-5x^2+x+2
zalisa [80]

The polynomial equation 2x^{3} - 5 x^{2} + x + 2 = 0 given that 2 is zero of

f(x) = 2x^{3} - 5 x^{2}  + x +2

According to the question,

f(x)  =2x^{3} - 5 x^{2} + x + 2

f(2)= 2(2)^{3} - 5 (2)^{2} + 2 + 2 \\= 2(8) - 5(4) +4\\= 16 - 20 + 4\\= -4 + 4\\= 0.

Hence, the polynomial equation 2x^{3} - 5 x^{2} + x + 2 = 0 given that 2 is zero of f(x)  =2x^{3} - 5 x^{2} + x + 2.

Learn more about polynomial equation here

brainly.com/question/1720316

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