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 #1
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What real value of t produces the smallest value of the quadratic t^2 -9t - 36 + 4 - 30t?  

 

Combine like terms                                    t2 – 39t – 32   

 

This is a parabola that opens upward.  

 

The smallest value will occur at the vertex.  That's the point where the curve turns around and heads back up. 

The first derivitive expresses the slope.  At the vertex, the slope equals zero.  Setting the slope equal to y:   

 

                                                                y  =  t2 – 39t – 32   

 

So the first derivitive is                             y'  =  2t – 39  

 

Set equal to zero                                      2t – 39  =  0 

 

                                                                   t  =  19.5    

 

The problem doesn't ask what   

the smallest value is, but you  

can find it by plugging 19.5  

into the original equation                    (19.5)2 – 39(19.5) – 32  

 

                                                            380.25 – 760.5 – 32  =  – 412.25   

 

Checked answer by pasting y=t^2–39t–32 into Desmos.  

 

After re-reading answer, I edited one word for terminology correction.  

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19.12.2023
 #2
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A sector of a circle is shown below.  The sector has an area of $60 \pi.$  What is the radius of the circle?  

 

I'm not sure what \pi means.  Does the backslant mean divided by pi, or is the backslant just another one of those useless symbols like all those dollar signs in every problem I look at.  I don't know about you, but I would use a forward slant to mean divided by.  I'm going to treat the backslant as meaningless.  

 

The shaded area is a fourth of the circle, indicated by the right angle in the center. 

 

                                                          Acircle  

                                                        –––––––  =  60 π  

                                                             4  

 

                                                           A  =  240 π  

 

                                                           A  =  π r2   

 

therefore                                       240 π  =  π r2     

 

                                                            r2  =  240  

 

                                                            r  =  sqrt(240)  

 

                                                            r  =  4 • sqrt(15)  

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18.12.2023