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coldgirl [10]
3 years ago
14

How to do a exponent with the negative base?

Mathematics
2 answers:
tatiyna3 years ago
3 0
Examples :
(-2)^3 = (-2)(-2)(-2) = - 8

(-3)^4 = (-3)(-3)(-3)(-3) = 81

(-1)^5 = (-1)(-1)(-1)(-1)(-1) = -1

** if the exponent is an even number, the answer will be positive. if the exponent is an odd number, then the answer will be negative

** but keep in mind....if the negative is not included with the base..
-(5^2) = - 25....but (-5^2) = 25
Sergeeva-Olga [200]3 years ago
3 0
Any number raised to a negative exponent could be written as follows:

a⁻ⁿ = 1/aⁿ

a⁻² = 1/a²

a⁻²³ = 1/a²³

a⁻ˣ = 1/aˣ
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(6, - 3, 1) and (8,9,-11) find the angle between the vector​
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Answer:

The angle between the two vectors is 84.813°.

Step-by-step explanation:

Statement is incomplete. Complete form is presented below:

<em>Let be (6,-3, 1) and (8, 9, -11) vector with same origin. Find the angle between the two vectors. </em>

Let \vec u = \langle 6, -3, 1 \rangle and \vec v = \langle 8,9,-11 \rangle, the angle between the two vectors is determined from definition of dot product:

\theta = \cos^{-1} \left(\frac{\vec u \,\bullet \,\vec v}{\|\vec u\|\cdot \|\vec v\|} \right) (1)

Where:

\vec u, \vec v - Vectors.

\|\vec u\|, \|\vec v\| - Norms of each vector.

Note: The norm of a vector in rectangular form can be determined by either the Pythagorean Theorem or definition of Dot Product.

If we know that \vec u = \langle 6,-3,1 \rangle and \vec v = \langle 8, 9,-11 \rangle, then the angle between the two vectors is:

\theta = \cos^{-1}\left[\frac{(6)\cdot (8) + (-3)\cdot (9) + (1)\cdot (-11)}{\sqrt{6^{2}+(-3)^{2}+1^{2}}\cdot \sqrt{8^{2}+9^{2}+(-11)^{2}}} \right]

\theta \approx 84.813^{\circ}

The angle between the two vectors is 84.813°.

6 0
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Help Me Please !! THX!
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Answer: decreases

Step-by-step explanation:

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How do I find the solution for -t/4=-3pi
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