A triangle can be right angled and scalene. true and false
2 answers:
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Answer:
Rational exponents are not defined when the denominator of the exponent in lowest terms is even and the base is negative.
Step-by-step explanation:
Considering the expression
Here:
- b = index
- a = exponent
- X = radicand
- √ = radical symbol
A rational exponent - an exponent that is a fraction - is the kind of way we may write a root.
If the denominator is an even number, it means we are talking about an even root like square root, 4th root, 6th root etc.
For example, think about squaring a number
-4 × -4 = 16, 4 × 4 = 16
It means any number when it get multiplied by itself an even number of times, it would always yield a positive number.
It is not possible to take the square root of a negative number as we can not yield a negative number when we square the number. In other words, there is no way we can multiply the same negative number twice and get a negative number. This is why is undefined.
Therefore, rational exponents are not defined when the denominator of the exponent in lowest terms is even and the base is negative.
Answer:
Step-by-step explanation:
The line integral with respect to arc length of the function f(x, y, z) = xy2 along the parametrized curve that is the line segment from (1, 1, 1) to (2, 2, 2) followed by the line segment from (2, 2, 2) to (−9, 6, 5) equals the sum of the line integral of f along each path separately.
Let
be the two paths.
Recall that if we parametrize a path C as with the parameter t varying on some interval [a,b], then the line integral with respect to arc length of a function f is
Given any two points P, Q we can parametrize the line segment from P to Q as
r(t) = tQ + (1-t)P with 0≤ t≤ 1
The parametrization of the line segment from (1,1,1) to (2,2,2) is
r(t) = t(2,2,2) + (1-t)(1,1,1) = (1+t, 1+t, 1+t)
r'(t) = (1,1,1)
and
The parametrization of the line segment from (2,2,2) to
(-9,6,5) is
r(t) = t(-9,6,5) + (1-t)(2,2,2) = (2-11t, 2+4t, 2+3t)
r'(t) = (-11,4,3)
and
Hence