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Simplify: 3x(4a - 5b) - 5x(3a - 2b). Combine: (2x-3)/(x-1). Solve: 2x/3 = 5 − 3x. Translate: 3 divided by the sum of a certain number and 8

Mathematics

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  1. Expand the first term by distributing $3x$

    3x(4a−5b)=12ax−15bx3x(4a - 5b) = 12ax - 15bx

  2. Expand the second term by distributing −5x-5x

    −5x(3a−2b)=−15ax+10bx-5x(3a - 2b) = -15ax + 10bx

  3. Combine the expanded terms

    (12ax−15bx)+(−15ax+10bx)(12ax - 15bx) + (-15ax + 10bx)

  4. Group like terms (axax terms and bxbx terms)

    (12ax−15ax)+(−15bx+10bx)(12ax - 15ax) + (-15bx + 10bx)

  5. Perform the subtraction for each group

    −3ax−5bx-3ax - 5bx

    The simplified expression is −3ax−5bx\boxed{-3ax - 5bx}.

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  6. Since the denominators are the same, combine the numerators over the common denominator

    2x−3x−1\frac{2x - 3}{x - 1}

  7. The expression cannot be simplified further.

    The expression as a single fraction is 2x−3x−1\boxed{\frac{2x - 3}{x - 1}}.

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  8. The equation is 2x3=5−3x\frac{2x}{3} = 5 - 3x.

  9. Multiply the entire equation by 3 to eliminate the denominator

    3×(2x3)=3×(5−3x)3 \times \left(\frac{2x}{3}\right) = 3 \times (5 - 3x) 2x=15−9x2x = 15 - 9x

  10. Add 9x9x to both sides to gather xx terms

    2x+9x=152x + 9x = 15 11x=1511x = 15

  11. Divide by 11 to solve for xx

    x=1511x = \frac{15}{11}

    The value of xx is 1511\boxed{\frac{15}{11}}.

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  12. Let the certain number be yy.

  13. Translate the problem into an equation

    "3 divided by the sum of a certain number and 8" is 3y+8\frac{3}{y+8}. "2 divided by the sum of that number and 3" is 2y+3\frac{2}{y+3}.

    3y+8=2y+3\frac{3}{y+8} = \frac{2}{y+3}

  14. Cross-multiply

    3(y+3)=2(y+8)3(y+3) = 2(y+8)

  15. Distribute on both sides

    3y+9=2y+163y + 9 = 2y + 16

  16. Subtract $2y$ from both sides

    3y−2y+9=163y - 2y + 9 = 16 y+9=16y + 9 = 16

  17. Subtract 9 from both sides

    y=16−9y = 16 - 9 y=7y = 7

    The number is 7\boxed{7}.

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  18. The equation is 103x−5=21\frac{10}{3x-5} = \frac{2}{1}.

  19. Cross-multiply

    10×1=2×(3x−5)10 \times 1 = 2 \times (3x - 5) 10=6x−1010 = 6x - 10

  20. Add 10 to both sides

    10+10=6x10 + 10 = 6x 20=6x20 = 6x

  21. Divide by 6 to solve for xx

    x=206x = \frac{20}{6}

  22. Simplify the fraction

    x=103x = \frac{10}{3}

    The value of xx is 103\boxed{\frac{10}{3}}.

    :

  23. Calculate the total length of the pitch.

    It is divided into 12 parts, each 25cm long.

    Total length=12×25 cm=300 cm\text{Total length} = 12 \times 25 \text{ cm} = 300 \text{ cm}

  24. Find how many parts, each 20cm long, can be obtained from the total length.

    Number of parts=Total lengthLength of each new part=300 cm20 cm\text{Number of parts} = \frac{\text{Total length}}{\text{Length of each new part}} = \frac{300 \text{ cm}}{20 \text{ cm}} Number of parts=15\text{Number of parts} = 15

    15 parts, each 20cm long, can be obtained from the pitch. 15 parts\boxed{15 \text{ parts}}.

    :

  25. Find the reciprocal of 0.65

    Reciprocal=10.65\text{Reciprocal} = \frac{1}{0.65}

  26. Calculate the value

    10.65≈1.53846...\frac{1}{0.65} \approx 1.53846...

  27. Round to 3 significant figures.

    The first three significant figures are 1, 5, 3. The next digit is 8, which is ≥5\ge 5, so round up the 3.

    1.53846...≈1.541.53846... \approx 1.54

    The reciprocal of 0.65 to 3 significant figures is 1.54\boxed{1.54}.

    :

  28. Calculate the fuel consumption rate (liters per km)

    Rate=100 liters250 km=1025 liters/km=25 liters/km=0.4 liters/km\text{Rate} = \frac{100 \text{ liters}}{250 \text{ km}} = \frac{10}{25} \text{ liters/km} = \frac{2}{5} \text{ liters/km} = 0.4 \text{ liters/km}

  29. Calculate the fuel needed for 150 km

    Fuel needed=Rate×Distance=0.4 liters/km×150 km\text{Fuel needed} = \text{Rate} \times \text{Distance} = 0.4 \text{ liters/km} \times 150 \text{ km} Fuel needed=60 liters\text{Fuel needed} = 60 \text{ liters}

    It will use 60 liters\boxed{60 \text{ liters}} of fuel.

    :

  30. The equation is P15=53\frac{P}{15} = \frac{5}{3}.

  31. To solve for PP, multiply both sides by 15

    P=53×15P = \frac{5}{3} \times 15

  32. Perform the multiplication

    P=5×153=5×5P = 5 \times \frac{15}{3} = 5 \times 5 P=25P = 25

    The value of PP is 25\boxed{25}.

    :

  33. Express ₦32.00 as a ratio of ₦72.00

    3272\frac{32}{72}

  34. Simplify the fraction by dividing both the numerator and the denominator by their greatest common divisor.

    - Both are divisible by 8:
    32÷872÷8=49\frac{32 \div 8}{72 \div 8} = \frac{4}{9}

    The ratio is 4:9\boxed{4:9}.

    :

  35. Calculate 15% of 4500kg

    Increase amount=15% of 4500 kg=15100×4500\text{Increase amount} = 15\% \text{ of } 4500 \text{ kg} = \frac{15}{100} \times 4500 Increase amount=15×45=675 kg\text{Increase amount} = 15 \times 45 = 675 \text{ kg}

  36. Add the increase amount to the original amount

    New amount=4500 kg+675 kg=5175 kg\text{New amount} = 4500 \text{ kg} + 675 \text{ kg} = 5175 \text{ kg}

    The increased amount is 5175 kg\boxed{5175 \text{ kg}}.

    :

  37. Find the prime factorization of each number

    - $9 = 3^2$
- $12 = 2^2 \times 3$
- $18 = 2 \times 3^2$

  38. Find the HCF (Highest Common Factor).

    This is the product of the lowest powers of common prime factors. The only common prime factor is 3, and its lowest power is 313^1.

    HCF(9,12,18)=3\text{HCF}(9, 12, 18) = 3

  39. Find the LCM (Least Common Multiple).

    This is the product of the highest powers of all prime factors present. - Highest power of 2 is 222^2. - Highest power of 3 is 323^2.

    LCM(9,12,18)=22×32=4×9=36\text{LCM}(9, 12, 18) = 2^2 \times 3^2 = 4 \times 9 = 36

  40. Find the product of the LCM and HCF

    Product=LCM×HCF=36×3=108\text{Product} = \text{LCM} \times \text{HCF} = 36 \times 3 = 108

    The product of the LCM and HCF is 108\boxed{108}.

    :

  41. Identify the given values for simple interest

    - Principal (P) = ₦30,000.00
- Time (T) = $3\frac{1}{2}$ years = 3.5 years
- Rate (R) = 5% per annum = 0.05

  42. Use the formula for simple interest

    Simple Interest (SI)=P×R×T100\text{Simple Interest (SI)} = \frac{P \times R \times T}{100}

  43. Substitute the values into the formula

    SI=30000×5×3.5100\text{SI} = \frac{30000 \times 5 \times 3.5}{100}

  44. Perform the multiplication in the numerator

    SI=30000×17.5100\text{SI} = \frac{30000 \times 17.5}{100} SI=525000100\text{SI} = \frac{525000}{100}

  45. Perform the division

    SI=₦5250.00\text{SI} = \text{₦}5250.00

    The simple interest is ₦5250.00\boxed{\text{₦}5250.00}.

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