With calculus you can show that the given line and the line tangent to the circle at the single intersection point are one and the same. But all you need to do is prove that there is only one intersection
What math will you be taking now? Will you have to start from scratch or just review? I went to sparknotes.com to review and look for tips on 3D geometry and sequences. Its pretty good for things on the test you may not have been exposed to in precalc class, but there probably arent enough...
|x| is either positive or negative, but not both at the same time. When taking the square root of both sides of an equation, you must take cases and usually come up with 2 possible solutions, a positive and negative answer
Thats exactly what it is! 2 equations with 2 unknowns, except your variables are f(2) and f(1/2). We have no clue what these can be, so you need to substitute or eliminate to solve for f(2). The hard part was noticing the trick of letting x=1/2. After that, its all algebra
Yes, thats the "range equation", assuming no air resistance. Notice that complementary angles result in the same range.
As for the original question, you dont need calculus for a parabola. Find the axis of symmetry and plug in to find the y coordinate of the vertex.
There is a relationship between b and the period of the function. Assuming this is a vertical position versus time function, the horizontal shift c is often called a phase shift and tells where equilibrium (y = 0 for a sine function) is in relation to the clock. With no phase shift, the function...
You need to use a sign graph because you dont know if the interval from 0 to 1 is positive or negative. If the question asked x2-x > 0 , your method of "just set them equal" would come up with the same interval. The correct answer is everything but 0 to 1, negative infinity to 0 or 1 to positive...
Its called indirect proof or proof by contradiction, and its a perfectly legitamate form of proof. Assume to the contrary (division is associative) and work until you reach something that cant be true
Along the same lines, its important to remember that there are just as many prime numbers and integers as there are natural numbers. Intuitively, youd think there would be more natural numbers, but the cardinality of all 3 sets is aleph0
Another thing to be careful of is when you take the square root of both sides of an equation.
\sqrt{x^2} doesn't equal x, it equals the absolute value of x
\sqrt{(-2)^2} = |-2| = 2
If you have:
x^2 = 4 and take the square root, you should get x = +/-2 because
\sqrt{x^2} = 2 = |x|
In the limit, only the highest degree matters. Because the numerator and denominator have the same degree, the function will have a limit (horizontal asymtote) approaching infinity. Divide the coefficient of the numerator by the coefficient of the denominator and you have all thats left when x...
This was taken from a math contest a few months ago.
Homework Statement
xx*yy=zz
find z if:
x=28 * 38
y=212 * 36
Homework Equations
Theres undoubtably some trick, but I have yet to find it
The Attempt at a Solution
Dont even think about calculator
I showed my math teacher, and...
Plug in 2. You get a nonzero number over 0, which means dividing a common factor wont help. The graph is asymtotic at x=2. You can show which infinity the function goes to with a sign graph. Find all important numbers (roots, asymtotes) and evaluate the function in between each number. The graph...
If you make the rather basic assumption that the question was posed over reals in an R^2 Cartesian plane, than imaginary numbers are not up for discussion. Furthermore, the concept of order (less than, greater than) dissolves in the complex number system, which makes delta-epsilon pretty hard to do
If youre asking about the limit of \sqrt{x}, you can only approach it from the right. Because the limit point is unapproachable from the left, the limit does not exist. Remember, its not a polynomial so you cant just plug it in
Then all thats left is to create a new limit as y approaches something. In this case its still 1 but you would need to take the 6th root of the original limit point
For example, if you wanted to do something like the limit as x approaches 8 of:
\displaystyle{\frac{-x^2+ 4x^\frac{4}{3}+...
Try solvng for x^2 and plugging that into one equation. Then plug in each y solution to get 2 corresponding x solutions. At the end, check to make sure all your answers work in the original equations
Well informally, after squaring both sides you have 128=2^(2m). But youve already identified that 128=2^7. So 7=2m
What youre doing is finding the log base 2 of both sides. Log base 2 of 128 = 7 because 2^7=128. Log base 2 of 2^(2m)=2m because 2^(2m)=well... 2^(2m)