Bosco

Find the coefficient of y^4 in the expansion of (2y-5+3y^2)^6.

That is way too much multiplication to do by hand.  

Fortunately, a useful tool has been provided for us, at  

https://www.symbolab.com/solver/expand-calculator

Per the "expand" tool on that page, the coefficient of y⁴ is 375.  

Would it be cheating, to use that website to avoid the drudgery of endless   

multiplications?  I do not consider it cheating to use any tool that has been  

made available.  It used to be slide rules, then calculators, now computers.   

_.

One ordered pair (a,b) satisfies the two equations ab^4 = 48 and ab = 72. What is the value of b in this ordered pair?    

To find b, consider                                                  ab⁴  =  48  

We will divide both sides by ab.  

Since ab=72, we will divide the left side  

by "ab" and the right side by its equal 72.  

                                                                              ab⁴         48  

                                                                             ——   =   ——  

                                                                              ab           72  

Note that ab⁴ = (ab) * (b³)  

Cancel ab out of the left side.  

Reduce 48/72 on the right side.  

                                                                               b³           2  

                                                                             ——   =   ——  

                                                                                1            3  

                                                                                 b   =   cube root of (2 / 3)  

_.

I drove to the beach at a rate of 40 miles per hour.  If I had driven at a rate of 50 miles per hour instead, then I would have arrived 45 minutes later.  How many miles did I drive?  

You mean 45 minutes earlier.  Obviously, if you drive faster, you get there faster.  

This problem makes use of

the following relationship:                   Distance = Velocity x Time 

                                                           D  =  V • T  

case 1                                                 D  =  (40) • (T)  

case 2                                                 D  =  (50) • (T – 45)  

Since the Distance, D, is the  

same for both cases, let's set       

the "V•T"s equal to each other.             (50)(T – 45)  =  (40)(T)  

                                                               50T – 2250  =  40T  

Subtract 40T from both sides                  10T – 2250  =  0  

Add 2250 to both sides                                       10T  =  2250  

Divide both sides by 10                                           T  =  225   (this is in minutes)  

Divide minutes by 60 to get hours          225 minutes  =  3.75 hours  

Plug this T back into original equation                     D  =  (40 mi/hr) • (3.75 hr)  =  150 miles  

_.

Ditto GA ... in spades.  I don't know what AoPS is, but I have noticed that the same, identical problems are stated over and over and over, literally identical, some with misspelled words, &/or incorrect punctuation, and everything else.  A notable example is the one where the question is how much later will someone arrive if he drives faster.  I've posted many answers by just copying/pasting my answer from an earlier instance. 

Don't get me started on the trolls who post one-word answers – like 66 – without explanation how the anwer was obtained.  They can't explain because they just pulled it out of their, er, I'll say their hat, but that's not what I mean.  

Hey Ginger!  

It's so great to hear from you.  I haven't been seeing postings of yours for quite a while, and I was afraid I never would again.  Thank you for that explanation about permutations.  I agree that Bosco the Chocolate Bear would have been pleased with my username.  Or is pleased, if he's watching from cat heaven. 

This is something funny.  It turns out that my username here is a cat I used to have, and my user name on eBay is a cat my grandma used to have.  I didn't plan it that way, and the coincidence just dawned on me a couple of days ago. 

Ron  

Will and Grace are canoeing on a lake.  Will rows at 50 meters per minute and Grace rows at 20 meters per minute. Will starts rowing at 2 p.m. from the west end of the lake, and Grace starts rowing from the east end of the lake at 2 p.m. If they always row directly towards each other, and the lake is 3800 meters across from the west side of the lake to the east side, at what time will the two meet?   

Will's rate is 50 m/min.  

Grace's rate is 20 m/min.  

They're rowing toward each other,  

so their rate of closure is 70 m/min.  

The lake is 3800 meters across.  

The time it will take them to meet is given by  

                                                                                    Distance  

                                                                            T  =  –––––––  

                                                                                       Rate  

                                                                                    3800 m  

                                                                            T  =  –––––––  

                                                                                    70 m/min  

The time it takes them to meet                             T  =  54.29 min  

Note that (0.29 min • 60 sec/min)  =  17 seconds

The time on the clock is                                             2:54:17 pm 

_.

Thank you, Alan.  I didn't know about that difference. 

_.

Find all values of x such that \sqrt{4x^2} + \sqrt{x^2} = 6 + 3x.  

The square root of 4x² is +2x.  

The square root of x² is +x,  

That gives us four propositions to address:  

1.    +2x + x  =  6 + 3x    combines to   0 = 6   dismiss as absurdum  

2,    +2x – x  =  6 + 3x    combines to   x = –3  

3.    –2x + x  =  6 + 3x    combines to   x = –1.5  

4.    –2x – x  =  6 + 3x    combines to   x = –1  

_.

Catherine rolls a standard 6-sided die six times.  If the product of her rolls is 2500 then how many different sequences of rolls could there have been?  (The order of the rolls matters.)  

There are two ways to obtain a product of 2500 using factors between 1 and 6 inclusive.  

One way is             5 • 5 • 5 • 5 • 2 • 2  

The other way is    5 • 5 • 5 • 5 • 4 • 1  

I admit I had to look that up.  

I searched on google for prime factorization of 2500.  

I don't know how to use permutations to mix those other numbers  

in there with all those fives, so I would have to use brute force.  

225555    522555    552255    555225    555522  

252555    525255    552525    555252   

255255    525525    552552   

255525    525552   

255552  

If I didn't miss any, there are 15 with 5's and 2's.  

There would be another 30 with 5's, a 4, and a 1. 

This is because there would be 15 with the 1 first, 

then another 15 with the 4 first.  I think that's right.  

I'm not going to write them all down.  If you want to  

write them all down, that would be okay with me. 

I think your teacher would be impressed. ;-)  

Anyway, the total number of sequences,  

according to my feeble calculations, seems to be 45 sequences.  

_.

I can't post pictures, so you'll have to visualize this along with me. 

The top and bottom corners of the square are secured to the top and bottom of the rectangle, respectively.  They can slide along the length of the rectangle, but they can't come loose.  

Push the square all the way to the left, so that its left corner is touching the left side of the rectangle.  You have just formed two triangles, whose area can never be covered by the square.  If we can figure out the area of those triangles, we'll be on our way to making an A in geometry.  

Notice that both triangles are right triangles, and they both have a hypotenuse that's 8 units long.  There's only a single configuration that accomplishes that, therefore the triangles are congruent.  

Take that bottom triangle and, hinging it on its sharpest-pointed angle, rotate it counterclockwise until its hypotenuse lies exactly on top of the hypotenuse of the top triangle.  You have just formed a rectangle, and we already know the length of its diagonal.  Here's the magic step.  Look it up if you don't believe me. 

The area of a rectangle is equal to half the square of its diagonal:  A = (1/2)(d²).  We can calculate the area of the small rectangle that you so cleverly formed, thus:  A = (1/2)(8²) = (1/2)(64) = 32.  Now double that because we have to do the same thing when we slide the square to the other end.

The question was "What is the area of the region inside the rectangle that the square can never cover?"  

The answer is, 64 square units.   Hmm, that's the same as the area of the square.  I wonder if it's a coincidence.  

~ edited to correct a couple of typos ~   

_.