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Find the image of a point P (3,–1) under the transformation defined by x' = x − 2y y' = 2x The figure on the grid below shows a triangular shaped object ABC and its image A'B'C'.

Mathematics

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Answer

Ksh. 57,618.15\text{Ksh. } 57,618.15

5, 6

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here is the solution to question 23 (b) and 24 (a):

b) Determine Mr. Kembo's net pay.

From the previous calculations (Question 23a), Mr. Kembo's gross pay (total taxable income) was Ksh. 76,000 and his PAYE was Ksh. 12,991.85.

  1. Calculate the NHIF contribution.

    The NHIF contribution is 2.75% of the gross pay.

    NHIF=0.0275×76,000\text{NHIF} = 0.0275 \times 76,000 NHIF=2,090 Ksh.\text{NHIF} = 2,090 \text{ Ksh.}

  2. Calculate the Housing levy.

    The Housing levy is 1.5% of the gross pay.

    Housing Levy=0.015×76,000\text{Housing Levy} = 0.015 \times 76,000 Housing Levy=1,140 Ksh.\text{Housing Levy} = 1,140 \text{ Ksh.}

  3. Calculate the total deductions.

    The total deductions include PAYE, NSSF, NHIF, and Housing levy.

    Total Deductions=PAYE+NSSF+NHIF+Housing Levy\text{Total Deductions} = \text{PAYE} + \text{NSSF} + \text{NHIF} + \text{Housing Levy} Total Deductions=12,991.85+2,160+2,090+1,140\text{Total Deductions} = 12,991.85 + 2,160 + 2,090 + 1,140 Total Deductions=18,381.85 Ksh.\text{Total Deductions} = 18,381.85 \text{ Ksh.}

  4. Calculate the net pay.

    Net pay is the gross pay minus total deductions.

    Net Pay=Gross Pay−Total Deductions\text{Net Pay} = \text{Gross Pay} - \text{Total Deductions} Net Pay=76,000−18,381.85\text{Net Pay} = 76,000 - 18,381.85 Net Pay=Ksh. 57,618.15\text{Net Pay} = \boxed{\text{Ksh. } 57,618.15}
    1. a) Find the image of a point P(3, -1) under the transformation defined by x′=x−2yx' = x - 2y and y′=2xy' = 2x.

    Given the point P(x,y)=P(3,−1)P(x, y) = P(3, -1), so x=3x = 3 and y=−1y = -1.

  5. Substitute the values of xx and yy into the equation for x′x'.

    x′=x−2yx' = x - 2y x′=3−2(−1)x' = 3 - 2(-1) x′=3+2x' = 3 + 2 x′=5x' = 5

  6. Substitute the value of xx into the equation for y′y'.

    y′=2xy' = 2x y′=2(3)y' = 2(3) y′=6y' = 6

  7. State the image point $P'(x', y')$.

    The image of point P(3,−1)P(3, -1) is P′(5,6)P'(\boxed{5, 6}).

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