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Calculate by using addition. Use the additive inverse of the integers in brackets.

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3. Calculate by using addition. Use the additive inverse of the integers in brackets.

Part a· 2 steps

  1. Use the additive inverse of −2-2, which is 2.

    4−(−2)=4+24 - (-2) = 4 + 2
  2. Perform the addition.

    4+2=64 + 2 = 6

    The answer is 6\boxed{6}.

Part b· 2 steps

  1. Use the additive inverse of +3+3, which is −3-3.

    8−(+3)=8+(−3)8 - (+3) = 8 + (-3)
  2. Perform the addition.

    8+(−3)=58 + (-3) = 5

    The answer is 5\boxed{5}.

Part c· 2 steps

  1. Use the additive inverse of +3+3, which is −3-3.

    −5−(+3)=−5+(−3)-5 - (+3) = -5 + (-3)
  2. Perform the addition.

    −5+(−3)=−8-5 + (-3) = -8

    The answer is −8\boxed{-8}.

Part d· 2 steps

  1. Use the additive inverse of −10-10, which is 10.

    9−(−10)=9+109 - (-10) = 9 + 10
  2. Perform the addition.

    9+10=199 + 10 = 19

    The answer is 19\boxed{19}.

Part e· 2 steps

  1. Use the additive inverse of −7-7, which is 7.

    8−(−7)=8+78 - (-7) = 8 + 7
  2. Perform the addition.

    8+7=158 + 7 = 15

    The answer is 15\boxed{15}.

Part f· 2 steps

  1. Use the additive inverse of +8+8, which is −8-8.

    −11−(+8)=−11+(−8)-11 - (+8) = -11 + (-8)
  2. Perform the addition.

    −11+(−8)=−19-11 + (-8) = -19

    The answer is −19\boxed{-19}.

Part g· 2 steps

  1. Use the additive inverse of +3+3, which is −3-3.

    −12−(+3)=−12+(−3)-12 - (+3) = -12 + (-3)
  2. Perform the addition.

    −12+(−3)=−15-12 + (-3) = -15

    The answer is −15\boxed{-15}.

Part h· 2 steps

  1. Use the additive inverse of −13-13, which is 13.

    15−(−13)=15+1315 - (-13) = 15 + 13
  2. Perform the addition.

    15+13=2815 + 13 = 28

    The answer is 28\boxed{28}.

Part h(i)· 17 steps

  1. Use the additive inverse of −15-15, which is 15.

    4−(−15)=4+154 - (-15) = 4 + 15
  2. Perform the addition.

    4+15=194 + 15 = 19

    The answer is 19\boxed{19}.

    j)

  3. Use the additive inverse of −13-13, which is 13.

    −13−(−13)=−13+13-13 - (-13) = -13 + 13
  4. Perform the addition.

    −13+13=0-13 + 13 = 0

    The answer is 0\boxed{0}.

    k)

  5. Use the additive inverse for each subtraction.

    The additive inverse of −1-1 is 1, and the additive inverse of −5-5 is 5.

    6−(−1)−(−5)=6+1+56 - (-1) - (-5) = 6 + 1 + 5
  6. Perform the addition.

    6+1+5=126 + 1 + 5 = 12

    The answer is 12\boxed{12}.

    l)

  7. Use the additive inverse for each subtraction.

    The additive inverse of −11-11 is 11, and the additive inverse of −8-8 is 8.

    19−(−11)−(−8)=19+11+819 - (-11) - (-8) = 19 + 11 + 8
  8. Perform the addition.

    19+11+8=3819 + 11 + 8 = 38

    The answer is 38\boxed{38}.

    4. A microwave oven increases the temperature of a piece of meat from −12∘C-12^\circ \text{C} to 67∘C67^\circ \text{C}. What is the change in temperature?

  9. Identify the initial and final temperatures.

    Initial temperature (TinitialT_{initial}) =−12∘C= -12^\circ \text{C}
    Final temperature (TfinalT_{final}) =67∘C= 67^\circ \text{C}

  10. Calculate the change in temperature using the formula ΔT=Tfinal−Tinitial\Delta T = T_{final} - T_{initial}.

    ΔT=67∘C−(−12∘C)\Delta T = 67^\circ \text{C} - (-12^\circ \text{C})
  11. Simplify the expression.

    Subtracting a negative number is equivalent to adding its positive counterpart.

    ΔT=67∘C+12∘C\Delta T = 67^\circ \text{C} + 12^\circ \text{C} ΔT=79∘C\Delta T = 79^\circ \text{C}

    The change in temperature is 79∘C\boxed{79^\circ \text{C}}.

    5. Mrs Mojapelo cooks chicken pieces and puts the dish in the freezer. The freezer lowers the temperature from 15∘C15^\circ \text{C} to −18∘C-18^\circ \text{C}. What is the temperature range?

  12. Identify the highest and lowest temperatures.

    Highest temperature =15∘C= 15^\circ \text{C}
    Lowest temperature =−18∘C= -18^\circ \text{C}

  13. Calculate the temperature range by subtracting the lowest temperature from the highest temperature.

    Range=15∘C−(−18∘C)\text{Range} = 15^\circ \text{C} - (-18^\circ \text{C})
  14. Simplify the expression.

    Range=15∘C+18∘C\text{Range} = 15^\circ \text{C} + 18^\circ \text{C} Range=33∘C\text{Range} = 33^\circ \text{C}

    The temperature range is 33∘C\boxed{33^\circ \text{C}}.

    6. Mr Sobekwa's bank statement shows an amount of −R500-\text{R}500. If he previously had R325\text{R}325 in his bank account, how much does he need to put back so that he once again has R325\text{R}325 in his account?

  15. Determine the current balance and the desired final balance.

    Current balance =−R500= -\text{R}500
    Desired final balance =R325= \text{R}325

  16. Calculate the amount needed to reach the desired final balance from the current balance.

    Amount needed=Desired final balance−Current balance\text{Amount needed} = \text{Desired final balance} - \text{Current balance} Amount needed=R325−(−R500)\text{Amount needed} = \text{R}325 - (-\text{R}500)
  17. Simplify the expression.

    Subtracting a negative amount is equivalent to adding a positive amount.

    Amount needed=R325+R500\text{Amount needed} = \text{R}325 + \text{R}500 Amount needed=R825\text{Amount needed} = \text{R}825

    Mr Sobekwa needs to put back R825\boxed{\text{R}825}.

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