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Step 1: Convert each fraction to a decimal to compare them.

Mathematics

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  1. Step 1: Convert each fraction to a decimal to compare them.
511≈0.4545\frac{5}{11} \approx 0.4545 37≈0.4286\frac{3}{7} \approx 0.4286 613≈0.4615\frac{6}{13} \approx 0.4615 25=0.4\frac{2}{5} = 0.4 49≈0.4444\frac{4}{9} \approx 0.4444

  1. Arrange the decimals in ascending order.

    0.4,0.4286,0.4444,0.4545,0.46150.4, 0.4286, 0.4444, 0.4545, 0.4615

  2. Write the original fractions in ascending order.

    25,37,49,511,613\boxed{\frac{2}{5}, \frac{3}{7}, \frac{4}{9}, \frac{5}{11}, \frac{6}{13}}
    1. Step 1: Identify the type of triangle. The markings indicate it is an isosceles triangle, meaning two sides are equal, and the base angles opposite those sides are also equal.

  3. Calculate the base angles of the isosceles triangle.

    The sum of angles in a triangle is 180∘180^\circ. Let the base angles be xx.

    48∘+x+x=180∘48^\circ + x + x = 180^\circ 48∘+2x=180∘48^\circ + 2x = 180^\circ 2x=180∘−48∘2x = 180^\circ - 48^\circ 2x=132∘2x = 132^\circ x=132∘2x = \frac{132^\circ}{2} x=66∘x = 66^\circ

  4. Determine the value of angle t∘t^\circ.

    Angle t∘t^\circ and the base angle xx form a linear pair on a straight line, so their sum is 180∘180^\circ.

    t∘+x=180∘t^\circ + x = 180^\circ t∘+66∘=180∘t^\circ + 66^\circ = 180^\circ t∘=180∘−66∘t^\circ = 180^\circ - 66^\circ t∘=114∘t^\circ = 114^\circ

    The value of the angle marked t∘t^\circ is 114∘\boxed{114^\circ}.

    1. Step 1: Use the area of a right-angled triangle formula to find the height. Area A=12×base×heightA = \frac{1}{2} \times \text{base} \times \text{height} Given A=96 m2A = 96 \text{ m}^2 and base b=12 mb = 12 \text{ m}.
    96 m2=12×12 m×h96 \text{ m}^2 = \frac{1}{2} \times 12 \text{ m} \times h 96 m2=6 m×h96 \text{ m}^2 = 6 \text{ m} \times h h=96 m26 mh = \frac{96 \text{ m}^2}{6 \text{ m}} h=16 mh = 16 \text{ m}

  5. Use the Pythagorean theorem to find the hypotenuse.

    Let the hypotenuse be cc.

    c2=b2+h2c^2 = b^2 + h^2 c2=(12 m)2+(16 m)2c^2 = (12 \text{ m})^2 + (16 \text{ m})^2 c2=144 m2+256 m2c^2 = 144 \text{ m}^2 + 256 \text{ m}^2 c2=400 m2c^2 = 400 \text{ m}^2 c=400 m2c = \sqrt{400 \text{ m}^2} c=20 mc = 20 \text{ m}

  6. Calculate the perimeter of the triangle.

    Perimeter P=base+height+hypotenuseP = \text{base} + \text{height} + \text{hypotenuse}

    P=12 m+16 m+20 mP = 12 \text{ m} + 16 \text{ m} + 20 \text{ m} P=48 mP = 48 \text{ m}

    The perimeter of the triangle is 48 m\boxed{48 \text{ m}}.

    1. Step 1: Identify the shape. The figure shows a semicircle with a right-angled triangle inscribed within it. The shaded region is the area of the semicircle minus the area of the triangle. The sides 42 cm42 \text{ cm} and 56 cm56 \text{ cm} are the legs of the right-angled triangle, and its hypotenuse is the diameter of the semicircle.

  7. Calculate the diameter of the semicircle using the Pythagorean theorem.

    d2=(42 cm)2+(56 cm)2d^2 = (42 \text{ cm})^2 + (56 \text{ cm})^2 d2=1764 cm2+3136 cm2d^2 = 1764 \text{ cm}^2 + 3136 \text{ cm}^2 d2=4900 cm2d^2 = 4900 \text{ cm}^2 d=4900 cm2d = \sqrt{4900 \text{ cm}^2} d=70 cmd = 70 \text{ cm}

  8. Calculate the radius of the semicircle.

    r=d2=70 cm2=35 cmr = \frac{d}{2} = \frac{70 \text{ cm}}{2} = 35 \text{ cm}

  9. Calculate the area of the semicircle.

    Use π=227\pi = \frac{22}{7}.

    Asemicircle=12πr2A_{semicircle} = \frac{1}{2} \pi r^2 Asemicircle=12×227×(35 cm)2A_{semicircle} = \frac{1}{2} \times \frac{22}{7} \times (35 \text{ cm})^2 Asemicircle=12×227×1225 cm2A_{semicircle} = \frac{1}{2} \times \frac{22}{7} \times 1225 \text{ cm}^2 Asemicircle=11×12257 cm2A_{semicircle} = 11 \times \frac{1225}{7} \text{ cm}^2 Asemicircle=11×175 cm2A_{semicircle} = 11 \times 175 \text{ cm}^2 Asemicircle=1925 cm2A_{semicircle} = 1925 \text{ cm}^2

  10. Calculate the area of the triangle.

    Atriangle=12×base×heightA_{triangle} = \frac{1}{2} \times \text{base} \times \text{height} Atriangle=12×42 cm×56 cmA_{triangle} = \frac{1}{2} \times 42 \text{ cm} \times 56 \text{ cm} Atriangle=21 cm×56 cmA_{triangle} = 21 \text{ cm} \times 56 \text{ cm} Atriangle=1176 cm2A_{triangle} = 1176 \text{ cm}^2

  11. Calculate the area of the shaded region.

    Ashaded=Asemicircle−AtriangleA_{shaded} = A_{semicircle} - A_{triangle} Ashaded=1925 cm2−1176 cm2A_{shaded} = 1925 \text{ cm}^2 - 1176 \text{ cm}^2 Ashaded=749 cm2A_{shaded} = 749 \text{ cm}^2

    The area of the shaded region is 749 cm2\boxed{749 \text{ cm}^2}.

    1. Step 1: List the number of slips of each color. Number of blue slips = 3 Number of red slips = 4 Number of green slips = 2 Number of yellow slips = 6

  12. Calculate the total number of slips.

    Total slips = 3+4+2+6=153 + 4 + 2 + 6 = 15

  13. Determine the number of red slips.

    Number of red slips = 4

  14. Calculate the probability of drawing a red slip.

    P(red)=Number of red slipsTotal number of slipsP(\text{red}) = \frac{\text{Number of red slips}}{\text{Total number of slips}} P(red)=415P(\text{red}) = \frac{4}{15}

    The probability of it being a red slip is 415\boxed{\frac{4}{15}}.

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