#page 716

[thestoopgallants]

[rapidpixeladventure]

Wonder aloud if the broken sandstorm longsword could be repaired.

{[PONDER]: [”Can someone repair that sandstorm longsword?”]}

Trying to not directly address Paige, you wonder aloud about the broken sword on the wall.

You can see Paige hesitate, but they swallow a lump in their throat and answer.

So the blade is cursed? Maybe that’s why Yean hasn’t repaired it. Paige still looks nervous so Cal chimes in to keep the conversation going.

Paige visibly relaxes as they reply. In fact this is the most relaxed you have seen them. They talk in an even tone and they barely stumble over their words.

Paige looks like they are almost... happy? It seems like you have found a subject they enjoy. Seran notices and tries to give Paige gentle support.

Paige waves their hands in the air, trying to dispel the compliment. But the faint blush and tiny hint of a smile implies that they are actually rather pleased with themselves.

You make a mental note of that, Paige likes metal-working and swords. Or at least you think they do.

[STATUS]

[QUESTS]

[MAP]

(discord server)

What Should We Do Now?

[abookblogofepicproportions]

You always have a choice. It's the one thing that no one can strip from us.

Luc, Origin by Jennifer L. Armentrout

[pwnthesat]

A school ordered $600 worth of lightbulbs. Some of the lightbulbs cost $1 each and the others cost $2 each. If the twice as many $1 bubls as $2 bulbs were ordered, how many lightbulbs were ordered altogether?

Think about it with small numbers first. If you bought twice as many $1 bulbs as $2 bulbs, then you buy 2 cheap bulbs for every expensive bulb. If you only buy 3 bulbs, what does it cost? $1 + $1 + $2 = $4. So every 3 bulbs cost $4. Are you with me on that?

Now, you know how much the school paid for all the bulbs. If you divide that by $4, you’ll get the number of small groups of 3 that we just talked about that the school bought. 600/4 = 150 groups of 3 bulbs.

Total bulbs: 150 × 3 = 450

There’s also algebra, of course. Say x = number of cheap bulbs and y = number of expensive bulbs.

x = 2yx + 2y = 600

Substitute and solve for y:

2y + 2y = 6004y = 600y = 150

Now find x:

x = 2(150)x = 300

So you have 300 cheap bulbs and 150 expensive bulbs—total bulbs 450, just like before. 

[pwnthesat]

Number 11 page 716 please and how could I salve those types of questions ??

I think the easiest way to go here is just try a few numbers until one works. For example, you might start by trying 6. When 6 is increased by 50% of itself (i.e. 3) you get 9. 9 isn’t between 10 and 20, so you need to go bigger.

Try 8. When 8 is increased by 50% of itself (i.e. 4) you get 12. That’s between 10 and 20, so it works. You’re done!

The algebra:

10 < n + 0.5n < 2010 < 1.5n < 206.6666… < n < 13.3333…

So just pick any even integer between 6.6 and 13.3 and you’re good.

[pwnthesat]

BB section4 #716 #11

Quick and dirty: just try a few things and see what happens. If n = 8, for example, then 50% of 8 is 4. 8 + 4 = 12, which is between 10 and 20. So n could be 8.

The algebra:

10 < n + 0.5n < 2010 < 1.5n < 206.6666… < n < 13.3333…

So n can be any EVEN integer between those numbers. 8, 10, and 12 work.