I think you mean zero to negative infinity is {x: x< or equal to 0}
There is no answer to this problem unless x is 0. For the suare root of 98x to be a real number, x has to be positive or zero. For the square root of -147x to be a real number, x has to be negative or zero. Seeing has x has to fit both requirements, the problem has an answer only if x is zero.
I will assume that this is sopposed to be integrated with respect to x. To make this problem easier, imagine that the integrand is x raised to the negative 3. The integral is 1/(-2x-2) plus some constant c.
e2x + ex = 0e2x = -exe2x / ex = -1ex = -1x = ln(-1)Actually, that's just a restatement of the original problem, but that'sas far as you can go with it, because there's no such thing as the log(or the ln) of a negative number. Your equation has no solution.
When you have a problem that you need to solve!
Your question is fairly vague, but I'm interpreting it as:What is the range of y=12cos(x)?Shortform:-1212(pi)/6-->6sqrt(3)~10.392(pi)/4-->6sqrt(2)~8.485(pi)/3-->6(pi)/2-->02(pi)/3-->-63(pi)/4-->-6sqrt(2)~-8.4855(pi)/6-->-6sqrt(3)~-10.392(pi)-->-12If you continue this, you'll notice that the values keep switching back and forth from 12 to -12 then back to 12, passing through all the values in between. This is to be expected, because if you look at the graph of cosine (as well as sine), it oscillates back and forth between two values, giving it a wave-like appearance. From this you can easily surmise that the maximum value that 12cos(x) will ever reach is 12 and the minimum it will ever reach is -12, giving you the range [-12,12].Conceptually, if you examine just the function cos(x), you realize that it oscillates back and forth between -1 and 1. So the function 12cos(x) will just take whatever results from cos(x) and multiply it by 12. Since the range of cos(x) is [-1,1], the range of 12cos(x) will just be 12 times the range of cos(x), [-12,12]. This works for any numerical amplitude modification of a sine or cosine function (putting a number in front of the function). The range of 5cos(x) would be [-5,5], the range of (pi)cos(x) would be [-(pi),(pi)], and so on for any real number.
The domain is the set of all possible x values, for this problem it would be negative infinity to positive infinity. The range is the set of all possible y values, for this problem it would be -2 too +2
It is still infinity. The problem is that (although it is easy to think of it this way) infinity is not a number. Infinity is, rather, the concept that something is boundless. Thus, "infinity + infinity" is a category error.
infinity/infinity if you are to wright this out, well... you can't, so therefore it is the longest.
1-infinity
Infinity added to anything is infinity (with the exception of -infinity, as it is an indeterminate form). Thus, infinity + 2 = infinity.The problem is that (although it is easy to think of it this way) infinity is not a number. Infinity is, rather, the concept that something is boundless.Thus, "infinity + 2" is a category error. (This is (supposed to be) a sum in maths, and infinity is not a number.)
Percent notation is a way of expressing a value as a fraction of 100. It is commonly used to represent percentages, where 100% is equal to the whole or 1. For example, 50% is equivalent to 0.5 or 50 out of 100.
Scientific notation is not a problem that needs to be "solved".
509,200,000 in Scientific Notation = 5.092 x 108
It truly depends on what the problem is, but normally a negative will change the overall answer of the sum: + and - = negative answer - and - = positive answer + and + = positive answer - and + = negative answer
It can be a problem to do with adding or subtracting or exponents.
55555
This problem can be expressed as, x - 7 ≤ -3. Adding 7 to both sides gives, x ≤ 4. Then the number can have any value in the range 4 to -∞ (negative infinity).