<p>If the value of 1/(t-1) is twice the value of 1/(t-2), what is the value of t?</p>
<p>I keep getting -5, but it doesn't fit...please help?</p>
<p>1/(t-1) = 2/(t-2)</p>
<p>(t-1) = (t-2)/2
2t-2 = t-2
t = 0</p>
<p>A sphere has radius 6 inches. What is the volume, in cubic inches, of the smallest cube that can contain the entire sphere?</p>
<p>I got 1728, but dunno if that’s right.
Thanks superjew. :)</p>
<p>(4/3)(pi)r^3</p>
<p>just plug in</p>
<p>and then you have to take a cube root and guess/check to find one that works</p>
<p>I figured it out with cross multiplication. dunno…why it didn’t make sense before…
1/(t-1) = 2/(t-2)
(t-2)= (2t-2)
0=t
`</p>
<p>I checked…and it’s actually 4 superjew…</p>
<p>1/(t-1) = 2/(t+2)
(t+2) = 2/(t+2)
t+2 = 2(t-1)
2= t-2
t=4</p>
<p>How does 4 work?</p>
<p>1(4-1) is not twice that of 1(4-2)</p>
<p>1\3 it not twice of 1\2.</p>
<p>How about this:</p>
<p>If 1/(t-1) is twice 1/(t-2) then 2/(t-1) = 1/(t-2).</p>
<p>Cross multipying gets 2t-4=t-1
t=3.
1/(3-1) = 1/2. 1/(3-2)=1</p>
<p>1/2*2 = 1</p>
<p>You posted the problem wrong. </p>
<p>Your first post says twice the value of t - 2 but you go on to say that it’s t + 2.</p>
<p>
</p>
<p>I think 1728 is right. The diameter of the sphere is 12 inches, so the cube’s side must be at least 12. 12^3 = 1728.</p>
<p>all the answers are correct.</p>
<ol>
<li>t= 0</li>
<li>12^3</li>
</ol>
<p>@superjew, you don’t have to find the volume of the sphere</p>
<p>yes, 1728 is right because the diameter of the sphere would be the length of one side of a cube.</p>
<p>12^3 = 1728</p>
<p>I guess collegeboard did a typo…because that was the problem on the online course…so yeah it is 0 :P</p>
<p>^I didn’t see a typo? I got 0 too from what you posted just like the other guy.</p>