{"id":969,"date":"2026-01-04T14:38:33","date_gmt":"2026-01-04T04:38:33","guid":{"rendered":"https:\/\/archive4ones.com\/2ndstudy\/?p=969"},"modified":"2026-01-04T16:40:33","modified_gmt":"2026-01-04T06:40:33","slug":"year11-math-2-2-2-warm-up-questions-shape-and-measurement-spatial-intelligence","status":"publish","type":"post","link":"https:\/\/archive4ones.com\/2ndstudy\/?p=969","title":{"rendered":"Year11 MATH 2-2-2 Warm-up Questions-Shape and Measurement (Spatial Intelligence)."},"content":{"rendered":"\n

<\/p>\n\n\n\n

To master Chapter 2 of General Mathematics, you must move beyond basic formulas and understand the proportional relationships<\/strong> between dimensions, area, and volume. This warm-up focuses on spatial reasoning, similarity, and the geometry of complex solids.<\/p>\n\n\n\n

Concepts and Skills Covered:<\/strong><\/h3>\n\n\n\n
    \n
  1. Spherical Geometry:<\/strong> Applying volume (V<\/mi>=<\/mo>4<\/mn>3<\/mn><\/mfrac>\u03c0<\/mi>r<\/mi>3<\/mn><\/msup><\/mrow>V = \\frac{4}{3}\\pi r^3<\/annotation><\/semantics><\/math>) and surface area (S<\/mi>A<\/mi>=<\/mo>4<\/mn>\u03c0<\/mi>r<\/mi>2<\/mn><\/msup><\/mrow>SA = 4\\pi r^2<\/annotation><\/semantics><\/math>) formulas.<\/li>\n\n\n\n
  2. Cones and Pyramids:<\/strong> Using V<\/mi>=<\/mo>1<\/mn>3<\/mn><\/mfrac>A<\/mi>h<\/mi><\/mrow>V = \\frac{1}{3}Ah<\/annotation><\/semantics><\/math> and understanding the role of slant height in surface area.<\/li>\n\n\n\n
  3. Similarity and Scaling:<\/strong> Understanding that if lengths scale by k<\/mi>k<\/annotation><\/semantics><\/math>, areas scale by k<\/mi>2<\/mn><\/msup>k^2<\/annotation><\/semantics><\/math> and volumes by k<\/mi>3<\/mn><\/msup>k^3<\/annotation><\/semantics><\/math>.<\/li>\n\n\n\n
  4. Composite Solids:<\/strong> Deconstructing complex shapes into standard geometric components.<\/li>\n\n\n\n
  5. Formula Manipulation:<\/strong> Solving for dimensions (like radius or height) when the total volume or area is known.<\/li>\n<\/ol>\n\n\n\n
    \n\n\n\n

    Warm-up: Shape and Measurement (Spatial Intelligence)<\/strong><\/p>\n\n\n\n

    Q1.<\/strong> A sphere has a radius of 3<\/mn> cm<\/mtext><\/mrow>3\\text{ cm}<\/annotation><\/semantics><\/math>. Calculate its volume in terms of \u03c0<\/mi>\\pi<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    A. 108\u03c0<\/em> cm3<\/p>\n\n\n\n

    B. 27\u03c0<\/em> cm3<\/p>\n\n\n\n

    C. 12\u03c0<\/em> cm3<\/p>\n\n\n\n

    D. 36\u03c0<\/em> cm3<\/p>\n\n\n\n

    Hint: The formula for the volume of a sphere is V<\/mi>=<\/mo>4<\/mn>3<\/mn><\/mfrac>\u03c0<\/mi>r<\/mi>3<\/mn><\/msup><\/mrow>V = \\frac{4}{3}\\pi r^3<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    Q2. <\/strong>A right cone has a radius of 5<\/mn> cm<\/mtext><\/mrow>5\\text{ cm}<\/annotation><\/semantics><\/math> and a slant height of 13<\/mn> cm<\/mtext><\/mrow>13\\text{ cm}<\/annotation><\/semantics><\/math>. What is its total surface area?<\/p>\n\n\n\n

    A. 65\u03c0<\/em> cm2<\/p>\n\n\n\n

    B. 115\u03c0<\/em> cm2<\/p>\n\n\n\n

    C. 155\u03c0<\/em> cm2<\/p>\n\n\n\n

    D. 90\u03c0<\/em> cm2<\/p>\n\n\n\n

    Hint: Add the area of the circular base to the curved surface area, \u03c0<\/em>r<\/em>2+\u03c0<\/em>r<\/em>l<\/em>.<\/p>\n\n\n\n

    Q3. <\/strong>A square-based pyramid has a base side length of 6<\/mn> m<\/mtext><\/mrow>6\\text{ m}<\/annotation><\/semantics><\/math> and a vertical height of 10<\/mn> m<\/mtext><\/mrow>10\\text{ m}<\/annotation><\/semantics><\/math>. Calculate its volume.<\/p>\n\n\n\n

    A. 180 m3<\/p>\n\n\n\n

    B. 120 m3<\/p>\n\n\n\n

    C. 60 m3<\/p>\n\n\n\n

    D. 360 m3<\/p>\n\n\n\n

    Hint: The volume of any pyramid is one-third the volume of a prism with the same base and height.<\/p>\n\n\n\n

    Q4. <\/strong>Two similar storage bins have a linear scale factor of k<\/mi>=<\/mo>4<\/mn><\/mrow>k = 4<\/annotation><\/semantics><\/math>. If the smaller bin holds 2<\/mn> L<\/mtext><\/mrow>2\\text{ L}<\/annotation><\/semantics><\/math> of grain, how much does the larger bin hold?<\/p>\n\n\n\n

    A. 8 L<\/p>\n\n\n\n

    B. 128 L<\/p>\n\n\n\n

    C. 16 L<\/p>\n\n\n\n

    D. 32 L<\/p>\n\n\n\n

    Hint: Remember that if lengths are multiplied by k<\/em>, the volume is multiplied by k<\/em>3.<\/p>\n\n\n\n

    Q5. <\/strong>The surface area of a sphere is 100<\/mn>\u03c0<\/mi> cm<\/mtext>2<\/mn><\/msup><\/mrow>100\\pi\\text{ cm}^2<\/annotation><\/semantics><\/math>. What is its radius?<\/p>\n\n\n\n

    A. 10 cm<\/p>\n\n\n\n

    B. 2.5 cm<\/p>\n\n\n\n

    C. 5 cm<\/p>\n\n\n\n

    D. 25 cm<\/p>\n\n\n\n

    Hint: Equate the given area to the formula 4\u03c0<\/em>r<\/em>2 and solve for r<\/em>.<\/p>\n\n\n\n

    Q6. <\/strong>Two similar triangles have areas of 10<\/mn> cm<\/mtext>2<\/mn><\/msup><\/mrow>10\\text{ cm}^2<\/annotation><\/semantics><\/math> and 90<\/mn> cm<\/mtext>2<\/mn><\/msup><\/mrow>90\\text{ cm}^2<\/annotation><\/semantics><\/math>. What is the linear scale factor k<\/mi>k<\/annotation><\/semantics><\/math> between them?<\/p>\n\n\n\n

    A. 3<\/p>\n\n\n\n

    B. 81<\/p>\n\n\n\n

    C. 4.5<\/p>\n\n\n\n

    D. 9<\/p>\n\n\n\n

    Hint: The ratio of the areas is equal to the square of the linear scale factor (k<\/em>2).<\/p>\n\n\n\n

    Q7. <\/strong> A cylinder and a cone have the same radius and the same vertical height. If the volume of the cone is 50<\/mn> cm<\/mtext>3<\/mn><\/msup><\/mrow>50\\text{ cm}^3<\/annotation><\/semantics><\/math>, what is the volume of the cylinder?<\/p>\n\n\n\n

    A. 150 cm3<\/p>\n\n\n\n

    B. 200 cm3<\/p>\n\n\n\n

    C. 100 cm3<\/p>\n\n\n\n

    D. 50 cm3<\/p>\n\n\n\n

    Hint: Compare the formulas V<\/em>cy<\/em>l<\/em>\u200b=\u03c0<\/em>r<\/em>2h<\/em> and V<\/em>co<\/em>n<\/em>e<\/em>\u200b=31\u200b\u03c0<\/em>r<\/em>2h<\/em>.<\/p>\n\n\n\n

    Q8. <\/strong>Calculate the total surface area of a closed hemisphere with a radius of 10<\/mn> cm<\/mtext><\/mrow>10\\text{ cm}<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    A. 300\u03c0<\/em> cm2<\/p>\n\n\n\n

    B. 400\u03c0<\/em> cm2<\/p>\n\n\n\n

    C. 200\u03c0<\/em> cm2<\/p>\n\n\n\n

    D. 150\u03c0<\/em> cm2<\/p>\n\n\n\n

    Hint: Don’t forget to include the area of the flat circular base (S<\/em>A<\/em>t<\/em>o<\/em>t<\/em>a<\/em>l<\/em>\u200b=3\u03c0<\/em>r<\/em>2).<\/p>\n\n\n\n

    Q9. <\/strong>If you double the radius of a cylinder while keeping its height the same, by what factor does the volume increase?<\/p>\n\n\n\n

    A. 16<\/p>\n\n\n\n

    B. 4<\/p>\n\n\n\n

    C. 2<\/p>\n\n\n\n

    D. 8<\/p>\n\n\n\n

    Hint: Look at the exponent of the radius in the cylinder volume formula.<\/p>\n\n\n\n

    Q10.<\/strong> A composite solid is made by placing a cone of height 4<\/mn> cm<\/mtext><\/mrow>4\\text{ cm}<\/annotation><\/semantics><\/math> on top of a cylinder of height 10<\/mn> cm<\/mtext><\/mrow>10\\text{ cm}<\/annotation><\/semantics><\/math>. Both have a radius of 3<\/mn> cm<\/mtext><\/mrow>3\\text{ cm}<\/annotation><\/semantics><\/math>. What is the total volume?<\/p>\n\n\n\n

    A. 102\u03c0<\/em> cm3<\/p>\n\n\n\n

    B. 94\u03c0<\/em> cm3<\/p>\n\n\n\n

    C. 126\u03c0<\/em> cm3<\/p>\n\n\n\n

    D. 114\u03c0<\/em> cm3<\/p>\n\n\n\n

    Hint: Calculate the volume of the cylinder and the cone separately, then add them together.<\/p>\n\n\n\n

    ****************************************************************************<\/p>\n\n\n\n

    Great work! Understanding how dimensions scale and interact in composite solids is the foundation for the more complex 3D modeling you’ll do in your PSMT. Keep up the momentum!<\/p>\n\n\n\n

    ****************************************************************************<\/p>\n\n\n\n

    \u4e00\u822c\u6570\u5b66\u306e\u7b2c2\u7ae0\u3092\u30de\u30b9\u30bf\u30fc\u3059\u308b\u306b\u306f\u3001\u57fa\u672c\u7684\u306a\u516c\u5f0f\u3092\u8d85\u3048\u3066\u3001\u5bf8\u6cd5\u3001\u9762\u7a4d\u3001\u4f53\u7a4d\u306e\u6bd4\u4f8b\u95a2\u4fc2\u3092\u7406\u89e3\u3059\u308b\u5fc5\u8981\u304c\u3042\u308a\u307e\u3059\u3002\u3053\u306e\u30a6\u30a9\u30fc\u30e0\u30a2\u30c3\u30d7\u3067\u306f\u3001\u7a7a\u9593\u7684\u63a8\u8ad6\u3001\u76f8\u4f3c\u3001\u8907\u96d1\u306a\u7acb\u4f53\u306e\u5e7e\u4f55\u5b66\u306b\u7126\u70b9\u3092\u5f53\u3066\u307e\u3059\u3002<\/p>\n\n\n\n

    \u30ab\u30d0\u30fc\u3055\u308c\u308b\u6982\u5ff5\u3068\u30b9\u30ad\u30eb\uff1a<\/strong><\/p>\n\n\n\n

    \u7403\u9762\u5e7e\u4f55\u5b66\uff1a\u4f53\u7a4d\uff08V<\/mi>=<\/mo>4<\/mn>3<\/mn><\/mfrac>\u03c0<\/mi>r<\/mi>3<\/mn><\/msup><\/mrow>V = \\frac{4}{3}\\pi r^3<\/annotation><\/semantics><\/math>\uff09\u3068\u8868\u9762\u7a4d\uff08S<\/mi>A<\/mi>=<\/mo>4<\/mn>\u03c0<\/mi>r<\/mi>2<\/mn><\/msup><\/mrow>SA = 4\\pi r^2<\/annotation><\/semantics><\/math>\uff09\u306e\u516c\u5f0f\u3092\u9069\u7528\u3057\u307e\u3059\u3002<\/p>\n\n\n\n

    \u5186\u9310\u3068\u30d4\u30e9\u30df\u30c3\u30c9\uff1aV<\/mi>=<\/mo>1<\/mn>3<\/mn><\/mfrac>A<\/mi>h<\/mi><\/mrow>V = \\frac{1}{3}Ah<\/annotation><\/semantics><\/math>\u3092\u4f7f\u7528\u3057\u3001\u8868\u9762\u7a4d\u306b\u304a\u3051\u308b\u659c\u9ad8\u306e\u5f79\u5272\u3092\u7406\u89e3\u3057\u307e\u3059\u3002<\/p>\n\n\n\n

    \u76f8\u4f3c\u3068\u30b9\u30b1\u30fc\u30ea\u30f3\u30b0\uff1a\u9577\u3055\u304ck<\/mi>k<\/annotation><\/semantics><\/math>\u3001\u9762\u7a4d\u304ck<\/mi>2<\/mn><\/msup>k^2<\/annotation><\/semantics><\/math>\u3001\u4f53\u7a4d\u304ck<\/mi>3<\/mn><\/msup>k^3<\/annotation><\/semantics><\/math>\u3067\u30b9\u30b1\u30fc\u30ea\u30f3\u30b0\u3055\u308c\u308b\u3053\u3068\u3092\u7406\u89e3\u3057\u307e\u3059\u3002<\/p>\n\n\n\n

    \u8907\u5408\u7acb\u4f53\uff1a\u8907\u96d1\u306a\u5f62\u72b6\u3092\u6a19\u6e96\u7684\u306a\u5e7e\u4f55\u5b66\u7684\u69cb\u6210\u8981\u7d20\u306b\u5206\u89e3\u3057\u307e\u3059\u3002<\/p>\n\n\n\n

    \u6570\u5f0f\u64cd\u4f5c\uff1a\u7dcf\u4f53\u7a4d\u307e\u305f\u306f\u7dcf\u9762\u7a4d\u304c\u65e2\u77e5\u306e\u5834\u5408\u3001\u5bf8\u6cd5\uff08\u534a\u5f84\u3084\u9ad8\u3055\u306a\u3069\uff09\u3092\u89e3\u304d\u307e\u3059\u3002<\/p>\n\n\n\n

    *******************<\/p>\n\n\n\n

    Right answers.<\/strong><\/p>\n\n\n\n

    Q1. D 36<\/mn>\u03c0<\/mi> cm<\/mtext>3<\/mn><\/msup><\/mrow>36\\pi\\text{ cm}^3<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Using V<\/mi>=<\/mo>4<\/mn>3<\/mn><\/mfrac>\u03c0<\/mi>r<\/mi>3<\/mn><\/msup><\/mrow>V = \\frac{4}{3}\\pi r^3<\/annotation><\/semantics><\/math>, we get 4<\/mn>3<\/mn><\/mfrac>\u00d7<\/mo>\u03c0<\/mi>\u00d7<\/mo>3<\/mn>3<\/mn><\/msup>=<\/mo>4<\/mn>3<\/mn><\/mfrac>\u00d7<\/mo>27<\/mn>\u03c0<\/mi>=<\/mo>36<\/mn>\u03c0<\/mi><\/mrow>\\frac{4}{3} \\times \\pi \\times 3^3 = \\frac{4}{3} \\times 27\\pi = 36\\pi<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    Q2. D 90<\/mn>\u03c0<\/mi> cm<\/mtext>2<\/mn><\/msup><\/mrow>90\\pi\\text{ cm}^2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Total surface area is the sum of the base (\u03c0<\/mi>r<\/mi>2<\/mn><\/msup>=<\/mo>25<\/mn>\u03c0<\/mi><\/mrow>\\pi r^2 = 25\\pi<\/annotation><\/semantics><\/math>) and the curved surface (\u03c0<\/mi>r<\/mi>l<\/mi>=<\/mo>65<\/mn>\u03c0<\/mi><\/mrow>\\pi rl = 65\\pi<\/annotation><\/semantics><\/math>).<\/p>\n\n\n\n

    Q3. B 120<\/mn> m<\/mtext>3<\/mn><\/msup><\/mrow>120\\text{ m}^3<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    The volume is 1<\/mn>3<\/mn><\/mfrac>\u00d7<\/mo>base area<\/mtext>\u00d7<\/mo>height<\/mtext>=<\/mo>1<\/mn>3<\/mn><\/mfrac>\u00d7<\/mo>36<\/mn>\u00d7<\/mo>10<\/mn>=<\/mo>120<\/mn><\/mrow>\\frac{1}{3} \\times \\text{base area} \\times \\text{height} = \\frac{1}{3} \\times 36 \\times 10 = 120<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    Q4. B 128<\/mn> L<\/mtext><\/mrow>128\\text{ L}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Volume scales by k<\/mi>3<\/mn><\/msup>k^3<\/annotation><\/semantics><\/math>. 4<\/mn>3<\/mn><\/msup>=<\/mo>64<\/mn><\/mrow>4^3 = 64<\/annotation><\/semantics><\/math>, so the new volume is 2<\/mn>\u00d7<\/mo>64<\/mn>=<\/mo>128<\/mn><\/mrow>2 \\times 64 = 128<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    Q5. C 5<\/mn> cm<\/mtext><\/mrow>5\\text{ cm}<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Setting 4<\/mn>\u03c0<\/mi>r<\/mi>2<\/mn><\/msup>=<\/mo>100<\/mn>\u03c0<\/mi><\/mrow>4\\pi r^2 = 100\\pi<\/annotation><\/semantics><\/math> leads to r<\/mi>2<\/mn><\/msup>=<\/mo>25<\/mn><\/mrow>r^2 = 25<\/annotation><\/semantics><\/math>, so r<\/mi>=<\/mo>5<\/mn><\/mrow>r = 5<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    Q6. A. 3<\/mn>3<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    The area scale factor is 90<\/mn>\/<\/mi>10<\/mn>=<\/mo>9<\/mn><\/mrow>90 \/ 10 = 9<\/annotation><\/semantics><\/math>. Since area scales by k<\/mi>2<\/mn><\/msup>k^2<\/annotation><\/semantics><\/math>, k<\/mi>=<\/mo>9<\/mn><\/msqrt>=<\/mo>3<\/mn><\/mrow>k = \\sqrt{9} = 3<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    Q7. A. 150<\/mn> cm<\/mtext>3<\/mn><\/msup><\/mrow>150\\text{ cm}^3<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    A cylinder’s volume is exactly three times that of a cone with the same dimensions.<\/p>\n\n\n\n

    Q8. A. 300<\/mn>\u03c0<\/mi> cm<\/mtext>2<\/mn><\/msup><\/mrow>300\\pi\\text{ cm}^2<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    The surface area is the sum of the curved hemisphere (2<\/mn>\u03c0<\/mi>r<\/mi>2<\/mn><\/msup>=<\/mo>200<\/mn>\u03c0<\/mi><\/mrow>2\\pi r^2 = 200\\pi<\/annotation><\/semantics><\/math>) and the flat circular base (\u03c0<\/mi>r<\/mi>2<\/mn><\/msup>=<\/mo>100<\/mn>\u03c0<\/mi><\/mrow>\\pi r^2 = 100\\pi<\/annotation><\/semantics><\/math>).<\/p>\n\n\n\n

    Q9. B. 4<\/mn>4<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    The volume formula \u03c0<\/mi>r<\/mi>2<\/mn><\/msup>h<\/mi><\/mrow>\\pi r^2 h<\/annotation><\/semantics><\/math> involves r<\/mi>2<\/mn><\/msup>r^2<\/annotation><\/semantics><\/math>. Doubling r<\/mi>r<\/annotation><\/semantics><\/math> means (<\/mo>2<\/mn>r<\/mi>)<\/mo>2<\/mn><\/msup>=<\/mo>4<\/mn>r<\/mi>2<\/mn><\/msup><\/mrow>(2r)^2 = 4r^2<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    Q10. A. 102<\/mn>\u03c0<\/mi> cm<\/mtext>3<\/mn><\/msup><\/mrow>102\\pi\\text{ cm}^3<\/annotation><\/semantics><\/math><\/p>\n\n\n\n

    Cylinder volume is 9<\/mn>\u03c0<\/mi>\u00d7<\/mo>10<\/mn>=<\/mo>90<\/mn>\u03c0<\/mi><\/mrow>9\\pi \\times 10 = 90\\pi<\/annotation><\/semantics><\/math>. Cone volume is 1<\/mn>3<\/mn><\/mfrac>\u00d7<\/mo>9<\/mn>\u03c0<\/mi>\u00d7<\/mo>4<\/mn>=<\/mo>12<\/mn>\u03c0<\/mi><\/mrow>\\frac{1}{3} \\times 9\\pi \\times 4 = 12\\pi<\/annotation><\/semantics><\/math>. Total is 102<\/mn>\u03c0<\/mi><\/mrow>102\\pi<\/annotation><\/semantics><\/math>.<\/p>\n\n\n\n

    \u7d20\u6674\u3089\u3057\u3044\u6210\u679c\u3067\u3059\uff01\u8907\u5408\u30bd\u30ea\u30c3\u30c9\u306b\u304a\u3051\u308b\u5bf8\u6cd5\u306e\u30b9\u30b1\u30fc\u30eb\u3068\u76f8\u4e92\u4f5c\u7528\u3092\u7406\u89e3\u3059\u308b\u3053\u3068\u306f\u3001PSMT\u3067\u884c\u3046\u3088\u308a\u8907\u96d1\u306a3D\u30e2\u30c7\u30ea\u30f3\u30b0\u306e\u57fa\u790e\u3068\u306a\u308a\u307e\u3059\u3002\u3053\u306e\u52e2\u3044\u3092\u7dad\u6301\u3057\u3066\u304f\u3060\u3055\u3044\uff01<\/p>\n\n\n\n

    <\/p>\n","protected":false},"excerpt":{"rendered":"

    To master Chapter 2 of General Mathematics, you must move beyond basic formulas and understand the proportional relationships between dimensions, area, and volume. This warm-up focuses on spatial reasoning, similarity, and the geometry of complex solids. Concepts and Skills Covered: Warm-up: Shape and Measurement (Spatial Intelligence) Q1. A sphere has a radius of 3 cm3\\text{ cm}. […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[8],"tags":[],"class_list":["post-969","post","type-post","status-publish","format-standard","hentry","category-math"],"_links":{"self":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts\/969","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=969"}],"version-history":[{"count":3,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts\/969\/revisions"}],"predecessor-version":[{"id":994,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=\/wp\/v2\/posts\/969\/revisions\/994"}],"wp:attachment":[{"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=969"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=969"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/archive4ones.com\/2ndstudy\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=969"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}