{"id":249163,"date":"2025-07-09T04:05:07","date_gmt":"2025-07-09T04:05:07","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=249163"},"modified":"2025-07-09T04:05:09","modified_gmt":"2025-07-09T04:05:09","slug":"the-diameter-d-of-a-sphere-is-15-8cm","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/09\/the-diameter-d-of-a-sphere-is-15-8cm\/","title":{"rendered":"The diameter, D, of a sphere is 15.8cm"},"content":{"rendered":"\n<p>The diameter, D, of a sphere is 15.8cm. Calculate the sphere&#8217;s volume, . Use the value 3.14 for ?, and round your answer to the nearest tenth. (Do not round any intermediate computations.)<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To calculate the volume of a sphere, we use the formula:V=43\u03c0r3V = \\frac{4}{3} \\pi r^3V=34\u200b\u03c0r3<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>VVV is the volume,<\/li>\n\n\n\n<li>\u03c0\\pi\u03c0 is approximately 3.14,<\/li>\n\n\n\n<li>rrr is the radius of the sphere.<\/li>\n<\/ul>\n\n\n\n<p>The problem provides the <strong>diameter<\/strong> of the sphere, D=15.8\u2009cmD = 15.8 \\, \\text{cm}D=15.8cm, so we need to find the <strong>radius<\/strong> rrr. The radius is half the diameter:r=D2=15.82=7.9\u2009cmr = \\frac{D}{2} = \\frac{15.8}{2} = 7.9 \\, \\text{cm}r=2D\u200b=215.8\u200b=7.9cm<\/p>\n\n\n\n<p>Now, we substitute the radius and the value of \u03c0\\pi\u03c0 into the volume formula:V=43\u00d73.14\u00d7(7.9)3V = \\frac{4}{3} \\times 3.14 \\times (7.9)^3V=34\u200b\u00d73.14\u00d7(7.9)3<\/p>\n\n\n\n<p>First, we calculate r3r^3r3:7.93=7.9\u00d77.9\u00d77.9=493.0397.9^3 = 7.9 \\times 7.9 \\times 7.9 = 493.0397.93=7.9\u00d77.9\u00d77.9=493.039<\/p>\n\n\n\n<p>Now, substitute r3=493.039r^3 = 493.039r3=493.039 into the volume formula:V=43\u00d73.14\u00d7493.039V = \\frac{4}{3} \\times 3.14 \\times 493.039V=34\u200b\u00d73.14\u00d7493.039V=43\u00d71545.218V = \\frac{4}{3} \\times 1545.218V=34\u200b\u00d71545.218V=2060.291\u2009cm3V = 2060.291 \\, \\text{cm}^3V=2060.291cm3<\/p>\n\n\n\n<p>Finally, rounding to the nearest tenth:V\u22482060.3\u2009cm3V \\approx 2060.3 \\, \\text{cm}^3V\u22482060.3cm3<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion:<\/h3>\n\n\n\n<p>The volume of the sphere is approximately <strong>2060.3 cm\u00b3<\/strong>.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-1097.jpeg\" alt=\"\" class=\"wp-image-249164\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>The diameter, D, of a sphere is 15.8cm. Calculate the sphere&#8217;s volume, . Use the value 3.14 for ?, and round your answer to the nearest tenth. (Do not round any intermediate computations.) The Correct Answer and Explanation is: To calculate the volume of a sphere, we use the formula:V=43\u03c0r3V = \\frac{4}{3} \\pi r^3V=34\u200b\u03c0r3 Where: [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-249163","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/249163","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=249163"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/249163\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=249163"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=249163"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=249163"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}