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Find the slope of the line segment that joins the points A(-2,-4) and B(3, 1)

Mathematics

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B

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13) To find the slope of the line segment joining points A(−2,−4)A(-2, -4) and B(3,1)B(3, 1), use the slope formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}.

  1. Identify the coordinates.

    (x1,y1)=(−2,−4)(x_1, y_1) = (-2, -4) (x2,y2)=(3,1)(x_2, y_2) = (3, 1)
  2. Substitute the values into the formula.

    m=1−(−4)3−(−2)m = \frac{1 - (-4)}{3 - (-2)} m=1+43+2m = \frac{1 + 4}{3 + 2} m=55m = \frac{5}{5} m=1m = 1

    The correct option is B.

    14) Two sets are equal if they contain exactly the same elements.

  3. List the elements of each set.

    X={a,b,1,2,3}X = \{a, b, 1, 2, 3\} Y={2,d,b,c}Y = \{2, d, b, c\} Z={1,b,c,d}Z = \{1, b, c, d\} W={b,2,1,3,a}W = \{b, 2, 1, 3, a\}
  4. Compare the sets.

    • Option a) Y≠ZY \neq Z because 2∈Y2 \in Y but 2∉Z2 \notin Z (and 1∈Z1 \in Z but 1∉Y1 \notin Y).
    • Option b) X≠ZX \neq Z because a∈Xa \in X but a∉Za \notin Z (and c,d∈Zc, d \in Z but c,d∉Xc, d \notin X).
    • Option c) Y≠WY \neq W because d,c∈Yd, c \in Y but d,c∉Wd, c \notin W (and 1,3,a∈W1, 3, a \in W but 1,3,a∉Y1, 3, a \notin Y).
    • Option d) X=WX = W because both sets contain the elements {a,b,1,2,3}\{a, b, 1, 2, 3\}. The order of elements does not matter for set equality. The correct option is D.

    15) Given tan⁡A=34\tan A = \frac{3}{4}. In a right-angled triangle, tan⁡A=oppositeadjacent\tan A = \frac{\text{opposite}}{\text{adjacent}}.

  5. Identify the opposite and adjacent sides.

    Opposite=3\text{Opposite} = 3 Adjacent=4\text{Adjacent} = 4
  6. Use the Pythagorean theorem to find the hypotenuse.

    Hypotenuse2=Opposite2+Adjacent2\text{Hypotenuse}^2 = \text{Opposite}^2 + \text{Adjacent}^2 Hypotenuse2=32+42\text{Hypotenuse}^2 = 3^2 + 4^2 Hypotenuse2=9+16\text{Hypotenuse}^2 = 9 + 16 Hypotenuse2=25\text{Hypotenuse}^2 = 25 Hypotenuse=25=5\text{Hypotenuse} = \sqrt{25} = 5
  7. Find sin⁡A\sin A.

    In a right-angled triangle, sin⁡A=oppositehypotenuse\sin A = \frac{\text{opposite}}{\text{hypotenuse}}.

    sin⁡A=35\sin A = \frac{3}{5}

    The correct option is D.

    16) To find the mean (average) score, sum all the scores and divide by the number of scores.

  8. Sum the scores.

    Sum=78+81+69+82+75=385\text{Sum} = 78 + 81 + 69 + 82 + 75 = 385
  9. Divide by the number of scores, which is 5.

    Mean=3855=77\text{Mean} = \frac{385}{5} = 77

    The correct option is B.

    17) To solve the proportion (r+1):7=20:28(r+1):7 = 20:28:

  10. Write the proportion as an equation of fractions.

    r+17=2028\frac{r+1}{7} = \frac{20}{28}
  11. Simplify the fraction on the right side.

    2028=5×47×4=57\frac{20}{28} = \frac{5 \times 4}{7 \times 4} = \frac{5}{7}
  12. Set the simplified fractions equal and solve for rr.

    r+17=57\frac{r+1}{7} = \frac{5}{7}

    Since the denominators are equal, the numerators must be equal.

    r+1=5r+1 = 5 r=5−1r = 5 - 1 r=4r = 4

    The correct option is A.

    18) To find the volume of a right cylinder, use the formula V=πr2hV = \pi r^2 h.

  13. Identify the given values.

    Base diameter =28= 28 cm, so radius r=282=14r = \frac{28}{2} = 14 cm.
    Height h=40h = 40 cm.
    Use π=227\pi = \frac{22}{7}.

  14. Substitute the values into the formula.

    V=227×(14 cm)2×40 cmV = \frac{22}{7} \times (14 \text{ cm})^2 \times 40 \text{ cm} V=227×196 cm2×40 cmV = \frac{22}{7} \times 196 \text{ cm}^2 \times 40 \text{ cm}
  15. Perform the calculation.

    V=22×1967×40 cm3V = 22 \times \frac{196}{7} \times 40 \text{ cm}^3 V=22×28×40 cm3V = 22 \times 28 \times 40 \text{ cm}^3 V=616×40 cm3V = 616 \times 40 \text{ cm}^3 V=24640 cm3V = 24640 \text{ cm}^3

    The correct option is C.

    19) This question requires data from a table or context not provided in the image. Therefore, it cannot be answered.

    20) This question requires data from a table or context not provided in the image. Therefore, it cannot be answered.

    21) To find the sum of 814498144_9 and 154291542_9:

  16. Add the numbers column by column from right to left, remembering that this is base 9 (digits 0-8).

    81449+15429\begin{array}{r} 8144_9 \\ + 1542_9 \\ \hline \end{array}
  17. Add the rightmost column (4+24+2).

    4+2=694 + 2 = 6_9
  18. Add the next column (4+44+4).

    4+4=894 + 4 = 8_9
  19. Add the next column (1+51+5).

    1+5=691 + 5 = 6_9
  20. Add the leftmost column (8+18+1).

    8+1=98 + 1 = 9

    Since 9 is equal to the base, write 0 and carry 1.

    9=1×9+0  ⟹  1099 = 1 \times 9 + 0 \implies 10_9

    So, the sum is 10686910686_9.
    The correct option is A.

    22) To find the prime factorization of 588:

  21. Divide by the smallest prime number, 2.

    588÷2=294588 \div 2 = 294
  22. Divide by 2 again.

    294÷2=147294 \div 2 = 147
  23. Divide by the next smallest prime number, 3.

    147÷3=49147 \div 3 = 49
  24. Divide by the next smallest prime number, 7.

    49÷7=749 \div 7 = 7
  25. The prime factors are $2, 2, 3, 7, 7$.

    588=22×31×72588 = 2^2 \times 3^1 \times 7^2

    The correct option is D.

    22) (This is a duplicate question number, but a new problem) To find Mr. Grabrah's commission:

  26. Identify the selling price and commission rate.

    Selling price = \43,770.00Commissionrate Commission rate= 3%$

  27. Calculate the commission.

    Commission=3% of $43,770.00\text{Commission} = 3\% \text{ of } \$43,770.00 Commission=0.03×$43,770.00\text{Commission} = 0.03 \times \$43,770.00 Commission=$1313.10\text{Commission} = \$1313.10 The correct option is C.

    23) To calculate the simple interest, use the formula I=PRTI = PRT.

  28. Identify the principal (PP), rate (RR), and time (TT).

    P=$37.00P = \$37.00

    R=2%=0.02R = 2\% = 0.02 T=9 yearsT = 9 \text{ years}
  29. Substitute the values into the formula.

    I=$37.00×0.02×9I = \$37.00 \times 0.02 \times 9

  30. Perform the calculation.

    I=$0.74×9I = \$0.74 \times 9 I=$6.66I = \$6.66 The correct option is C.

    24) The logarithm property for the product of two numbers is log⁡b(XY)=log⁡bX+log⁡bY\log_b (XY) = \log_b X + \log_b Y.

  31. Apply the product rule for logarithms.

    log⁡bMN=log⁡bM+log⁡bN\log_b MN = \log_b M + \log_b N

    The correct option is A.

    25) To evaluate the expression P(1+rt)P(1+rt) with the given values:

  32. Substitute P=2P=2, r=−5r=-5, and t=3t=3 into the expression.

    P(1+rt)=2(1+(−5)(3))P(1+rt) = 2(1 + (-5)(3))
  33. Perform the multiplication inside the parenthesis first.

    2(1+(−15))2(1 + (-15))
  34. Perform the addition inside the parenthesis.

    2(1−15)2(1 - 15) 2(−14)2(-14)
  35. Perform the final multiplication.

    2×(−14)=−282 \times (-14) = -28

    The final answer is −28\boxed{-28}.

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