How do you solve these math word problems?

Question

I have 2 questions on my hw that are difficult for me to solve. Can someone tell me how to solve them?? Here they are:

1)The heat loss of a glass window varies jointly as the window's area and the difference between the outside and inside temperatures. A window 3 feet wide by 6 feet long loses 1200 Btu per hour when the temperature outside is 20 degrees colder than the temperature inside. Find the heat loss through a glass window that is 6 feet wide by 9 feet long when the temperature outside is 10 degrees colder that the temperature inside.........................................................................................................................


2) Sound intensity varies inversely as the square of the distance from the sound source. If you are in a movie theater and you change your seat to one that is twice as far from the speakers, how does the new sound intensity compare to that of your original seat?



Thanks in advance for the help I really appreciate i

Answer 1

1.) The first window's given area is 18 sq ft, and the temperatue difference is 20 degrees. This will let us find a constant for this equation we'll create:
Heat loss = c * a * ΔT
1200 = c * 18 * 20 = 360c
c = 1200 / 360 = 3 1/3

For the second window:
Heat loss = c * a * ΔT
Heat loss = 3 1/3 * (6 * 9) * 10
Heat loss = 1800 BTU

2.) If sound intensity varies inversely with the square of the distance, you can represent this as:
I = 1/d^2

If we double the distance:
I = 1/(2d)^2
I = 1/4d^2
I = 1/4d^2

so by doubling the distance, we have 1/4th the intensity.

Answer 2

The way I solve word problems is to find the numbers involved and the read the last sentence
to find out what I am suppose to do with those
numbers.
1. 3x6/1200BTU per hour/20 degrees
6x9/?BTU per hour/10 degrees
Find the heat loss. Even though the window is larger the temp is warmer so the BTU's should be less.
2. The further away from the speakers you are
the less intense the sound will be.
I hope this helps a little.

Answer 3

answer: one million + sqrt(2) miles enable s1 be the cost of the grasp ans s2 the cost of the prepare. The grasp and prepare commute in opposite guidelines until the grasp reaches the back of the prepare at time t1 at a distance d from the grasp's commencing element. So, d = s1 * t1. The prepare's rear is d miles from the grasp's commencing element, so the prepare's front must be at a distance of one million - d miles from that element, which became additionally the prepare's commencing element. The prepare has for this reason traveled one million - d miles, so one million - d = s2 * t1. If we divide d by making use of one million - d we get d/(one million-d) = s1/s2. as quickly as the grasp reaches the tip of the prepare, his horse does a piroette and he rides to the front of the prepare, achieving it because it hits the only million mile mark. The prepare might have traveled distance d and the grasp distance one million + d. So s1 * t2 = one million + d and s2 * t2 = d. branch provides us s1/s2 = (one million+d)/d. d/(one million-d) = s1/s2 = (one million+d)/d d^2 = one million - d^2 2d^2 = one million d^2 = one million/2 d = sqrt(2)/2 The grasp rode distance d, then one million + d for a whole of one million + 2d = one million + sqrt(2)