<h3>
Answer: 1</h3>
where x is nonzero
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Explanation:
We'll use two rules here
- (a^b)^c = a^(b*c) ... multiply exponents
- a^b*a^c = a^(b+c) ... add exponents
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The portion [ x^(a-b) ]^(a+b) would turn into x^[ (a-b)(a+b) ] after using the first rule shown above. That turns into x^(a^2 - b^2) after using the difference of squares rule.
Similarly, the second portion turns into x^(b^2-c^2) and the third part becomes x^(c^2-a^2)
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After applying rule 1 to each of the three pieces, we will have 3 bases of x with the exponents of (a^2-b^2), (b^2-c^2) and (c^2-a^2)
Add up those exponents (using rule 2 above) and we get
(a^2-b^2)+(b^2-c^2)+(c^2-a^2)
a^2-b^2+b^2-c^2+c^2-a^2
(a^2-a^2) + (-b^2+b^2) + (-c^2+c^2)
0a^2 + 0b^2 + 0c^2
0+0+0
0
All three exponents add to 0. As long as x is nonzero, then x^0 = 1
Answer:
Step-by-step explanation:
2y=11-3
2y=8
Y=8/2
Y=4
Answer:
Step-by-step explanation:
The question to be solved is the following :
Suppose that a and b are any n-vectors. Show that we can always find a scalar γ so that (a − γb) ⊥ b, and that γ is unique if 
. Recall that given two vectors a,b a⊥ b if and only if 
where 
is the dot product defined in 
. Suposse that 
. We want to find γ such that 
. Given that the dot product can be distributed and that it is linear, the following equation is obtained
Recall that 
are both real numbers, so by solving the value of γ, we get that
By construction, this γ is unique if 
, since if there was a 
such that 
, then
