Solve the math problems in the attached image.

Question image

Understand the Problem

The image presents three math problems. The first requires calculating the value of an expression involving fractions. The second involves understanding bearings in geometry. The third involves identifying the largest and smallest numbers from a set of numbers expressed in scientific notation.

Answer

6. $\frac{7}{12}$ 7. $320^\circ$ 8. (a) $1.36 \times 10^{6}$ (b) $5.21 \times 10^{-3}$
Answer for screen readers
  1. $\frac{7}{12}$
  2. $320^\circ$
  3. (a) $1.36 \times 10^{6}$ (b) $5.21 \times 10^{-3}$

Steps to Solve

  1. Problem 6: Find a common denominator

To subtract fractions, we need a common denominator. The least common multiple of 3 and 12 is 12. Convert $\frac{2}{3}$ to a fraction with a denominator of 12: $$ \frac{2}{3} = \frac{2 \times 4}{3 \times 4} = \frac{8}{12} $$

  1. Problem 6: Subtract the fractions

Now that both fractions have the same denominator, we can subtract them: $$ \frac{8}{12} - \frac{1}{12} = \frac{8-1}{12} = \frac{7}{12} $$

  1. Problem 6: Simplify the fraction

Check if the fraction $\frac{7}{12}$ can be simplified. Since 7 is a prime number and 7 does not divide 12, the fraction is already in its simplest form.

  1. Problem 7: Calculate the bearing of $B$ from $A$

The bearing of $A$ from $B$ is $140^\circ$. This means that to go from $B$ to $A$, you move $140^\circ$ clockwise from North. The bearing of $B$ from $A$ will be $180^\circ$ different if the bearing of $A$ from $B$ is less than $180^\circ$, or $180^\circ$ less if the bearing from $A$ to $B$ is greater than $180^\circ$. In this case, the bearing of $A$ from $B$ is $140^\circ$, which is less than $180^\circ$, therefore we have $$ 140^\circ + 180^\circ = 320^\circ $$ Alternatively, visualize the two towns $A$ and $B$. The angle at $B$ is $140^\circ$ clockwise from the North line. The angle at $A$ is $180^\circ + 140^\circ = 320^\circ$ clockwise from the North line.

  1. Problem 8(a): Identify the largest number

We are given the following numbers in standard form: $3.4 \times 10^{-1}$, $1.36 \times 10^{6}$, $7.9 \times 10^{0}$, $2.4 \times 10^{5}$, $5.21 \times 10^{-3}$, $4.3 \times 10^{-2}$

To find the largest number, we look for the number with the largest exponent of 10. The exponents are -1, 6, 0, 5, -3, -2. The largest exponent is 6. Therefore, the largest number is $1.36 \times 10^{6}$.

  1. Problem 8(b): Identify the smallest number

To find the smallest number, we look for the number with the smallest exponent of 10. The exponents are -1, 6, 0, 5, -3, -2. The smallest exponent is -3. Therefore, the smallest number is $5.21 \times 10^{-3}$.

  1. $\frac{7}{12}$
  2. $320^\circ$
  3. (a) $1.36 \times 10^{6}$ (b) $5.21 \times 10^{-3}$

More Information

Bearings are always measured clockwise from North. They are usually given as three-figure numbers e.g. $030^\circ$.

Tips

  • In problem 6, a common mistake is not finding a common denominator before subtracting the fractions. Another mistake is forgetting to simplify the fraction to its simplest form at the end.
  • In problem 7, a lot of people may think that since the bearing of A from B is $140^\circ$ the bearing of B from A is $140^\circ$ as well, but this is wrong.
  • In problem 8, be very careful to compare the exponents when the numbers are in standard form to determine which number is the largest and smallest.

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