Summarize the following:
This will be summer in the northern hemisphere and winter in the southern hemisphere. These months are the best to view the Milky Way because it is not as close to the sun. You may be able to see parts of the Milky Way as early as March or as late as August. Between November and February, however, it is not visible. The hours just after sunset and before sunrise are still very bright. Wait at least two hours after the sun has set before you go out stargazing. You can use an almanac or a weather website to learn when sunset and sunrise will happen on any particular day. Use this to plan your trip. Light from buildings, streets, and cars can block your view. Try to find a rural site that is far away from any towns, houses, or major roads. Since the Milky Way appears in the southern sky, you should travel south of any major cities. If you do this, the light from the city will not interfere with your viewing of the Milky Way. Nature reserves, mountains, deserts, and other unsettled areas are great places to view the Milky Way. To find dark areas, you can use a light pollution map, such as this one: http://www.youcanseethemilkyway.com/light-pollution/. You may not be able to see the Milky Way if the moon is too bright or if clouds are blocking the sky. Before going out to see the Milky Way, pick a clear night with either a new moon or a crescent moon. Most weather services will tell you both how many clouds there will be as well as what phase the moon is in. Several apps, such as Luna Solaria or Moon Phase Plus, will tell you what phase the moon is in. During this time, do not use a flashlight, phone, or another source of light. Your eyes need time to adjust to the darkness before you can see stars.

summary: Search for the Milky Way between June and August. Start stargazing two hours after sunset and before dawn. Find a location that has no light pollution. Pick a moonless and cloudless night. Let your eyes adapt for 20 minutes.

Summarize the following:
Congratulations! You've differentiated your equation implicitly not an easy task for first-timers! Using this equation to find the slope (dy/dx) for any (x, y) point is as simple as plugging in the x and y values for your point into the right side of the equation, then solving for (dy/dx). For example, let's say that we want to find the slope at the point (3, -4) for our example equation above. To do this, we would substitute 3 for x and -4 for y, solving as follows: (dy/dx) = (-2y2 - 2x + 5)/(2(2xy + y + 4) (dy/dx) = (-2(-4)2 - 2(3) + 5)/(2(2(3)(-4) + (-4) + 4) (dy/dx) = (-2(16) - 6 + 5)/(2(2(3)(-4)) (dy/dx) = (-32) - 6 + 5)/(2(2(-12)) (dy/dx) = (-33)/(2(2(-12)) (dy/dx) = (-33)/(-48) = 3/48, or 0.6875. The chain rule is an important piece of knowledge to have when dealing with calculus problems (including implicit differentiation problems). The chain rule states that for a function F(x) which can be written as (f o g)(x), the derivative of F(x) is equal to f'(g(x))g'(x). For difficult implicit differentiation problems, this means that it's possible to differentiate different individual "pieces" of the equation, then piece together the result. As a simple example, let's say that we need to find the derivative of sin(3x2 + x) as part of a larger implicit differentiation problem for the equation sin(3x2 + x) + y3 = 0. If we think of sin(3x2 + x) as "f(x)" and 3x2 + x as "g(x)", we can find the differentiation as follows: f'(g(x))g'(x) (sin(3x2 + x))' (3x2 + x)' cos(3x2 + x) (6x + 1) (6x + 1)cos(3x2 + x) Though it's not common in basic calculus, some advanced applications may require the implicit differentiation of more than two variables. For each extra variable, you'll need to find an extra derivative with respect to x. For instance, if you're working with x, y, and z, you'll need to find both (dz/dy) and (dz/dx). We can do this by differentiating the equation with respect x twice the first time, we'll insert a (dz/dx) every time we differentiate a term with z, and the second time, we'll insert a (dz/dy) every time we differentiate a z. After this, it's just a matter of solving for (dz/dx) and (dz/dy). For example, let's say that we're trying to differentiate x3z2 - 5xy5z = x2 + y3. First, let's differentiate with respect to x and insert (dz/dx). Don't forget to apply the product rule where appropriate! x3z2 - 5xy5z = x2 + y3 3x2z2 + 2x3z(dz/dx) - 5y5z - 5xy5(dz/dx) = 2x 3x2z2 + (2x3z - 5xy5)(dz/dx) - 5y5z = 2x (2x3z - 5xy5)(dz/dx) = 2x - 3x2z2 + 5y5z (dz/dx) = (2x - 3x2z2 + 5y5z)/(2x3z - 5xy5) Now, let's do the same for (dz/dy) x3z2 - 5xy5z = x2 + y3 2x3z(dz/dy) - 25xy4z - 5xy5(dz/dy) = 3y2 (2x3z - 5xy5)(dz/dy) = 3y2 + 25xy4z (dz/dy) = (3y2 + 25xy4z)/(2x3z - 5xy5)

summary: Plug in (x, y) values to find (dy/dx) for any point. Use the chain rule for functions-within-functions. For equations with x, y, and z variables, find (dz/dx) and (dz/dy).

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Don't use water balloons, which are too thin and weak for this purpose, and will easily burst. For a typical palm-sized stress ball, you'll need approximately 160 to 240 grams (5.6 to 8.5oz) of filling. Any of these will do: For a firm stress ball, use flour, baking soda, or cornstarch (a white powder called cornflour in the Commonwealth). For a looser stress ball, use dry rice lentils, small beans, or split peas, or fine play sand from a hardware store. Mix a small amount of dried rice into flour for something in between. This is also more durable than flour alone.
summary: Gather an uninflated balloon. Choose a filling.