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What is SheepShaver?
SheepShaver is a MacOS run-time environment for BeOS and Linux that allows you to run classic MacOS applications inside the BeOS/Linux multitasking environment. This means that both BeOS/Linux and MacOS applications can run at the same time (usually in a window on the BeOS/Linux desktop) and data can be exchanged between them. If you are using a PowerPC-based system, applications will run at native speed (i.e. with no emulation involved). There is also a built-in PowerPC emulator for non-PowerPC systems.
SheepShaver is distributed under the terms of the GNU General Public License (GPL). However, you still need a copy of MacOS and a PowerMac ROM image to use SheepShaver. If you're planning to run SheepShaver on a PowerMac, you probably already have these two items.
Supported systems
SheepShaver runs with varying degree of functionality on the following systems:
- Unix with X11 (Linux i386/x86_64/ppc, NetBSD 2.x, FreeBSD 3.x)
- Mac OS X (PowerPC and Intel)
- Windows NT/2000/XP
- BeOS R4/R5 (PowerPC)
Some of SheepShaver's features
- Runs MacOS 7.5.2 thru 9.0.4. MacOS X as a guest is not supported.
- Color video display
- CD quality sound output
- Access to floppy disks, CD-ROMs and HFS(+) partitions on hard disks
- Easy file exchange with the host OS via a 'Host Directory Tree' icon on the Mac desktop
- Internet and LAN networking via Ethernet
- Serial drivers
- SCSI Manager (old-style) emulation
Hi Shelly,
I like this problem a lot partly because there are so many ways to solve it. If your son knows some algebra he can solve it using the algebraic technique used on the algebra.com web site.
This problem can however be presented to a student as soon as he or she learns to do arithmetic with decimals. I think this is wonderful problem to give to a group of such students and have them work on a joint solution. I am going to use the numeric values in the problem referred to above and refer to the three chickens as a rooster, a hen and a chick with the rooster the heaviest and the chick the lightest.
The rooster and the chick weigh 8.5 kg and the hen and chick weigh 6.1 kg. Thus the rooster weighs 8.5 - 6.1 = 2.4 kg more than the chicken. But the rooster and the hen together weigh 10.6 kg. Hence, on the scale with the rooster and the hen replace the rooster by an identical hen and a 2.4 kg weight. The scale will still read 10.6 kg. Remove the 2.4 kg weight and the scale reads 10.6 - 2.4 = 8.2 kg. Thus two chickens weigh 8.2 kg. What does one chicken weigh?
A third solution comes from an observation. What does 10.6 + 8.5 + 6.1 = 25.2 kg represent in terms of roosters, hens and chicks?
I hope this helps,
Penny
I think I might know what you mean, and if so, it's not a ratio problem but a problem in linear algebra.
Are the data in a form like:
X and Y together weigh A
Y and Z together weigh B
Z and X together weigh C ?
This gives rise to the linear equation system
x + y = A
y + z = B
x + z = C
Qq 10.6.8 Software
This isn't in triangular form, so has no obvious 'weak spot.' But you can (for instance) subtract the second equation from the first to get a second equation in x and z; then add this to the third to eliminate z.
Qq 10.6.8 Download
Good Hunting!
RD