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<p>The cost to produce<code class='latex inline'>x</code> units of a product can be modelled by the function <code class='latex inline'>c(x)=\sqrt{x}+500</code>. </p><p><strong>(a)</strong> State and interpret the domain and range of the cost function.</p>
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<p>Find the area of each figure.</p><img src="/qimages/34393" />
Similar Question 3
<p>The formula to determine the distance <code class='latex inline'>d</code> in miles that an object can be seen on a clear day on the surface of a body of water is <code class='latex inline'>d=1.4\sqrt{h}</code>􏳎, where <code class='latex inline'>h</code> is the height in feet of the viewer’s eyes above the surface of the water.</p><p>The observation deck of a lighthouse stands 120 feet above the ocean surface. Can the lighthouse keeper see a boat that is 17 miles from the lighthouse? Explain.</p>
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Learning Path
L1 Quick Intro to Factoring Trinomial with Leading a
L2 Introduction to Factoring ax^2+bx+c
L3 Factoring ax^2+bx+c, ex1
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<p>On Earth, the time, <code class='latex inline'>t</code>, in seconds, taken for an object to fall from a height, <code class='latex inline'>h</code> , in metres, to the ground is given by the formula <code class='latex inline'>t(h) = \sqrt{\frac{h}{4.9}}</code>. On the moon, the formula changes to <code class='latex inline'>t(h)= \sqrt{\frac{h}{1.8}}</code>.</p><p>Graph both relations on the same set of axes. Compare the graphs and describe any similarities or differences.</p>
<p>On Earth, the time, <code class='latex inline'>t</code>, in seconds, taken for an object to fall from a height, <code class='latex inline'>h</code> , in metres, to the ground is given by the formula <code class='latex inline'>t(h) = \sqrt{\frac{h}{4.9}}</code>. On the moon, the formula changes to <code class='latex inline'>t(h)= \sqrt{\frac{h}{1.8}}</code>.</p><p>Determine the difference between the time it takes for an object to fall from a height of 25 m on Earth and the time it takes on the moon. Justify your answer.</p>
<p>When an object is dropped from a height, the time it takes to reach the ground is a function of the height from which it was dropped. An equation for this function is <code class='latex inline'>t(h)=\sqrt{\frac{h}{4.9}}</code>, where <code class='latex inline'>h</code> is in metres and <code class='latex inline'>t</code> is in seconds. </p><p>(a) Describe the domain and range of the function.</p><p>(b) Sketch the graph by applying a transformation to the graph of <code class='latex inline'>t(h)=\sqrt{h}</code>.</p>
<p>Find the area of each figure.</p><img src="/qimages/34393" />
<p>The formula to determine the distance <code class='latex inline'>d</code> in miles that an object can be seen on a clear day on the surface of a body of water is <code class='latex inline'>d=1.4\sqrt{h}</code>􏳎, where <code class='latex inline'>h</code> is the height in feet of the viewer’s eyes above the surface of the water.</p><p>The observation deck of a lighthouse stands 120 feet above the ocean surface. Can the lighthouse keeper see a boat that is 17 miles from the lighthouse? Explain.</p>
<p>Think About a Plan The radius of a spherical balloon can be expressed as <code class='latex inline'>\displaystyle r=\sqrt[3]{\frac{3 V}{4 \pi}} </code> inches, where <code class='latex inline'>\displaystyle r </code> is the radius and <code class='latex inline'>\displaystyle V </code> is the volume of the balloon in cubic inches. If air is pumped to inflate the balloon from 500 cubic inches to 800 cubic inches, by how many inches has the radius of the balloon increased?</p> <ul> <li><p>What was the radius of the balloon originally?</p></li> <li><p>What was the radius after inflating the balloon to 800 cubic inches?</p></li> <li><p>How can you use the two radii to find the amount of increase?</p></li> </ul>
<p>CCSS SENSE-MAKING A women&#39;s regulation-sized basketball is slightly smaller than a men&#39;s basketball. The radius <code class='latex inline'>\displaystyle r </code> of the ball that holds <code class='latex inline'>\displaystyle V </code> cubic units of air is <code class='latex inline'>\displaystyle \left(\frac{3 V}{4 \pi}\right)^{\frac{1}{3}} </code>.</p><p>a. Find the radius of a women&#39;s basketball.</p><p>b. Find the radius of a men&#39;s basketball.</p>
<p>GEOMETRY The radius <code class='latex inline'>\displaystyle r </code> of a sphere with volume <code class='latex inline'>\displaystyle V </code> is given by <code class='latex inline'>\displaystyle r=\left(\frac{3 V}{4 \pi}\right)^{\frac{1}{3}} </code>. Find the radius of a ball with a volume of <code class='latex inline'>\displaystyle 77 \mathrm{~cm}^{3} </code>.</p>
<p>Think About a Plan A square picture on the front page of a newspaper occupies an area of <code class='latex inline'>\displaystyle 24 \mathrm{in.}^{2} </code>. What is the length of each side of the picture? Write your answer as a radical in simplified form.</p> <ul> <li><p>How can you find the side length of a square if you know the area?</p></li> <li><p>What property can you use to write your answer in simplified form?</p></li> </ul>
<p>The formula to determine the distance <code class='latex inline'>d</code> in miles that an object can be seen on a clear day on the surface of a body of water is <code class='latex inline'>d=1.4\sqrt{h}</code>􏳎, where <code class='latex inline'>h</code> is the height in feet of the viewer’s eyes above the surface of the water.</p><p>Dillan and Marissa are parasailing while on vacation. Marissa is 135 feet above the ocean while Dillan is 85 feet above the ocean. How much farther can Marissa see than Dillan?</p>
<p>The initial velocity, <code class='latex inline'>v</code>, in kilometres per hour, of a skidding car can be determined from the length, <code class='latex inline'>d</code>, in metres, of the skid mark made by using the relation <code class='latex inline'>v(d) = 12.6\sqrt{d} + 8</code>.</p><p>(a) Determine the domain and range of the relation.</p><p>(b) Graph the relation.</p><p>(c) Is the relation a function? Justify your answer.</p>
<p>An object is dropped from a height of 80 m above the surface of a planet. On Earth, the relation <code class='latex inline'>t(h) = \sqrt{\dfrac{80-h}{4.9}}</code> represents the time, t, in seconds, when the object is at a height of h metres above the ground. On Jupiter, the relation is <code class='latex inline'>t(h) = \sqrt{\dfrac{80-h}{12.9}}</code>.</p><p>a) Express each relation using mapping notation.</p><p>b) Determine the domain and range of each relation.</p><p>c) ls each relation a function? Explain.</p><p>d) Determine the times when the object is 10 m above the ground on Earth and on Jupiter.</p>
<p>GARDENING If the area <code class='latex inline'>\displaystyle A </code> of a square is known, then the lengths of its sides <code class='latex inline'>\displaystyle \ell </code> can be computed using <code class='latex inline'>\displaystyle \ell=A^{\frac{1}{2}} </code>. You have purchased a <code class='latex inline'>\displaystyle 169 \mathrm{ft}^{2} </code> share in a community garden for the season. What is the length of one side of your square garden?</p>
<p>Find the area of each figure.</p><img src="/qimages/34394" />
<p>On Earth, the time, <code class='latex inline'>t</code>, in seconds, taken for an object to fall from a height, <code class='latex inline'>h</code> , in metres, to the ground is given by the formula <code class='latex inline'>t(h) = \sqrt{\frac{h}{4.9}}</code>. On the moon, the formula changes to <code class='latex inline'>t(h)= \sqrt{\frac{h}{1.8}}</code>.</p><p>Express each relation using mapping notation.</p>
<p>On Earth, the time, <code class='latex inline'>t</code>, in seconds, taken for an object to fall from a height, <code class='latex inline'>h</code> , in metres, to the ground is given by the formula <code class='latex inline'>t(h) = \sqrt{\frac{h}{4.9}}</code>. On the moon, the formula changes to <code class='latex inline'>t(h)= \sqrt{\frac{h}{1.8}}</code>.</p><p>Is each relation a function? Explain.</p>
<p>The cost to produce<code class='latex inline'>x</code> units of a product can be modelled by the function <code class='latex inline'>c(x)=\sqrt{x}+500</code>. </p><p><strong>(b)</strong> Suppose that the cost to make 10 prototype units is to be included in the cost. Write a new function representing the cost of this product. </p>
<p>The cost to produce<code class='latex inline'>x</code> units of a product can be modelled by the function <code class='latex inline'>c(x)=\sqrt{x}+500</code>. </p><p><strong>(a)</strong> State and interpret the domain and range of the cost function.</p>
<p>The cost to produce<code class='latex inline'>x</code> units of a product can be modelled by the function <code class='latex inline'>c(x)=\sqrt{x}+500</code>. </p><p><strong>(c)</strong> What type of transformation does the change in part (b) represent?</p><p><strong>(d)</strong> How does the transformation in part (b) affect the domain and range?</p>
<p>The formula to determine the distance <code class='latex inline'>d</code> in miles that an object can be seen on a clear day on the surface of a body of water is <code class='latex inline'>d=1.4\sqrt{h}</code>􏳎, where <code class='latex inline'>h</code> is the height in feet of the viewer’s eyes above the surface of the water.</p><p>A charter plane is used to fly tourists on a sightseeing trip along the coast of North Carolina. If the plane flies at an altitude of 1500 feet, how far can the tourists see?</p>
<p>Sports The bases in a softball diamond are located at the corners of a <code class='latex inline'>\displaystyle 3600-\mathrm{ft}^{2} </code> square. How far is a throw from second base to home plate?</p><img src="/qimages/60369" />
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