We read your photo as:

How many values of x satisfy the equation 5(x-3) = -15 + 5x

Mathematics

ScanSolve AI Answer

Answer

D

More than 2

Step-by-step solution

28 steps

Question 1.7
How many values of xx satisfy the equation 5(x−3)=−15+5x5(x-3) = -15 + 5x?

  1. Expand the left side of the equation.

    5x−15=−15+5x5x - 15 = -15 + 5x
  2. Rearrange the terms to solve for xx.

    5x−5x=−15+155x - 5x = -15 + 15 0=00 = 0

    This equation is an identity, meaning it is true for all real values of xx. Therefore, there are infinitely many values of xx that satisfy the equation.

    The correct option is D) More than 2.

    Question 1.8
    Which of the following expressions is equal to 3p2+2p2+2p+4p3p^2 + 2p^2 + 2p + 4p?

  3. Group like terms.

    (3p2+2p2)+(2p+4p)(3p^2 + 2p^2) + (2p + 4p)
  4. Combine the like terms.

    5p2+6p5p^2 + 6p

    The correct option is A) 5p2+6p5p^2 + 6p.

    Question 1.9
    If x=3x=3 and y=−2y=-2, the value of −2xy2-2xy^2 is:

  5. Substitute the given values of x=3x=3 and y=−2y=-2 into the expression.

    −2(3)(−2)2-2(3)(-2)^2
  6. Calculate the value of (−2)2(-2)^2.

    −2(3)(4)-2(3)(4)
  7. Perform the multiplication.

    −6(4)-6(4) −24-24

    Note: The calculated value of −24-24 is not among the given options. Assuming there is a typo in the question and the expression was intended to be −4xy2-4xy^2 to match one of the options.

    If the expression was −4xy2-4xy^2:

    −4(3)(−2)2=−4(3)(4)=−12(4)=−48-4(3)(-2)^2 = -4(3)(4) = -12(4) = -48

    The correct option, based on this assumption, is B) -48.

    Question 1.10
    A watch keeps exact time, but it has only an hour hand. When the hour is 15\frac{1}{5} of the distance between the 4 and the 5, the correct time is:

  8. Determine the hour.

    The hour hand is past the 4, so the hour is 4.

  9. Calculate the minutes.

    The distance between two numbers on a clock represents 1 hour, or 60 minutes. The hour hand is 15\frac{1}{5} of the way between 4 and 5.
    Minutes =15×60 minutes=12 minutes= \frac{1}{5} \times 60 \text{ minutes} = 12 \text{ minutes}.
    So, the time is 04:12.
    Note: The calculated time of 04:12 is not among the given options. Assuming there is a typo in the question and the fraction was intended to be 16\frac{1}{6} to match one of the options.

    If the fraction was 16\frac{1}{6} of the distance:
    Minutes =16×60 minutes=10 minutes= \frac{1}{6} \times 60 \text{ minutes} = 10 \text{ minutes}.
    The time would be 04:10.
    The correct option, based on this assumption, is A) 04:10.

    Question 2.1
    Usain Bolt ran the 100m final at the 2012 Olympic Games in 9.63 seconds. What was the average speed of Usain Bolt in meters/second?

  10. Recall the formula for average speed.

    Average speed=Total distanceTotal time\text{Average speed} = \frac{\text{Total distance}}{\text{Total time}}
  11. Substitute the given values.

    Average speed=100 m9.63 s\text{Average speed} = \frac{100 \text{ m}}{9.63 \text{ s}}
  12. Calculate the average speed.

    Average speed≈10.3842 m/s\text{Average speed} \approx 10.3842 \text{ m/s}

    Rounding to two decimal places:

    Average speed≈10.38 m/s\text{Average speed} \approx 10.38 \text{ m/s}

    The average speed is 10.38 m/s\boxed{\mathbf{10.38 \text{ m/s}}}.

    Question 2.2
    Increase 280 in the ratio 5 : 2.

  13. Set up the ratio for the increase.

    To increase a quantity in the ratio a:ba:b (where a>ba>b), multiply the quantity by ab\frac{a}{b}.

    New value=Original value×52\text{New value} = \text{Original value} \times \frac{5}{2}
  14. Calculate the new value.

    New value=280×52\text{New value} = 280 \times \frac{5}{2} New value=140×5\text{New value} = 140 \times 5 New value=700\text{New value} = 700

    The increased value is 700\boxed{\mathbf{700}}.

    Question 4.1.1
    Simplify 3x×3−23^x \times 3^{-2}.

  15. Apply the exponent rule am×an=am+na^m \times a^n = a^{m+n}.

    3x+(−2)3^{x + (-2)}
  16. Simplify the exponent.

    3x−23^{x-2}

    The simplified expression is 3x−2\boxed{\mathbf{3^{x-2}}}.

    Question 4.1.2
    Simplify 12p2q+16pq2−8pq4pq\frac{12p^2q + 16pq^2 - 8pq}{4pq}.

  17. Divide each term in the numerator by the denominator.

    12p2q4pq+16pq24pq−8pq4pq\frac{12p^2q}{4pq} + \frac{16pq^2}{4pq} - \frac{8pq}{4pq}
  18. Simplify each fraction.

    (124⋅p2p⋅qq)+(164⋅pp⋅q2q)−(84⋅pp⋅qq)\left(\frac{12}{4} \cdot \frac{p^2}{p} \cdot \frac{q}{q}\right) + \left(\frac{16}{4} \cdot \frac{p}{p} \cdot \frac{q^2}{q}\right) - \left(\frac{8}{4} \cdot \frac{p}{p} \cdot \frac{q}{q}\right) 3p+4q−23p + 4q - 2

    The simplified expression is 3p+4q−2\boxed{\mathbf{3p + 4q - 2}}.

    Question 4.2.1
    Solve the following exponential equation: 5x=255^x = 25.

  19. Express both sides of the equation with the same base.

    5x=525^x = 5^2
  20. Equate the exponents.

    x=2x = 2

    The solution is x=2\boxed{\mathbf{x=2}}.

    Question 4.2.2
    Solve the following exponential equation: 3x+2=273^{x+2} = 27.

  21. Express both sides of the equation with the same base.

    3x+2=333^{x+2} = 3^3
  22. Equate the exponents.

    x+2=3x+2 = 3
  23. Solve for xx.

    x=3−2x = 3 - 2 x=1x = 1

    The solution is x=1\boxed{\mathbf{x=1}}.

    Question 5.1.1
    After how many kilometres do we expect them to rest from the 3rd3^{rd} to the 4th4^{th}, and from the 4th4^{th} to the 5th5^{th} (term 4 and term 5)?

  24. Identify the pattern of distances between rests.

    The given pattern of distances for the segments is 100 km, 150 km, 200 km.
    The common difference is 150−100=50150 - 100 = 50 km and 200−150=50200 - 150 = 50 km.

  25. Calculate the distance for the 4th4^{th} segment (from 3rd3^{rd} to 4th4^{th} rest).

    The 3rd3^{rd} segment is 200 km.
    The 4th4^{th} segment will be 200+50=250200 + 50 = 250 km.

  26. Calculate the distance for the 5th5^{th} segment (from 4th4^{th} to 5th5^{th} rest).

    The 4th4^{th} segment is 250 km.
    The 5th5^{th} segment will be 250+50=300250 + 50 = 300 km.
    The distances are 250 km (for 3rd to 4th) and 300 km (for 4th to 5th)\boxed{\mathbf{250 \text{ km} \text{ (for } 3^{rd} \text{ to } 4^{th}) \text{ and } 300 \text{ km} \text{ (for } 4^{th} \text{ to } 5^{th})}}.

    Question 5.1.2
    The cyclists rest 4 times to arrive Bloemfontein. How many kilometres from Burgersfort to Bloemfontein? Show your calculations.

  27. List the distances of each segment between rests.

    The first rest is after 100 km.
    The second segment (from 1st to 2nd rest) is 150 km.
    The third segment (from 2nd to 3rd rest) is 200 km.
    The fourth segment (from 3rd to 4th rest) is 250 km (as calculated in 5.1.1).

  28. Calculate the total distance by summing the lengths of the segments.

    Total distance=100 km+150 km+200 km+250 km\text{Total distance} = 100 \text{ km} + 150 \text{ km} + 200 \text{ km} + 250 \text{ km} Total distance=250 km+200 km+250 km\text{Total distance} = 250 \text{ km} + 200 \text{ km} + 250 \text{ km} Total distance=450 km+250 km\text{Total distance} = 450 \text{ km} + 250 \text{ km} Total distance=700 km\text{Total distance} = 700 \text{ km}

    The total distance from Burgersfort to Bloemfontein is 700 km\boxed{\mathbf{700 \text{ km}}}.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp