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First right answer gets 10pts, tricky algebra?
I'm just testing yahoo answers to see who here actually knows what they're talking about (at least in higher algebra)
Here's a link to a moderately hard advanced algebra problem, it can be tricky if you think you know what your talking about when you don't. It basically focuses on people's understanding of exponents.
http://img146.imageshack.us/img146/9964/...
I'll post an answer soon enough, Go for it!
I just don't like people posting answers to other questions when they don't understand the concepts
I'm getting an answer from this question even if its not the mathematical answer
yes mina! you're on the right track
all good, but keep going! scratch the last step and try getting a more simplified answer
bun, good idea, just you can't combine it like that
check mina's way and follow it until that last step then keep going
9 Answers
- Anonymous1 decade agoFavorite Answer
Question::: : 2(4^K + 4^K)^2 ------------- (1)
Solution :::: :
= 2( 2*4K)^2
= 2( 2^2*4^2K)
= 2^3 * 4^2K
= 2^3 * (2^2)^2K
= 2^3 * 2^4K
= 2^(3+4K)---------------------(2) ANSWER
Proof ::::
In eqn 1 Substitute K=2------------ 2048
In eqn 2 Substitue K=2------------ 2048
Tats it…..
- 1 decade ago
2(4^(k)+4^(k))^(2)
Squaring an expression is the same as multiplying the expression by itself 2 times.
(2*(4^(k)+4^(k))(4^(k)+4^(k)))
Multiply each term in the first group by each term in the second group using the FOIL method. FOIL stands for First Outer Inner Last, and is a method of multiplying two binomials. First, multiply the first two terms in each binomial group. Next, multiply the outer terms in each group, followed by the inner terms. Finally, multiply the last two terms in each group.
(2(4^(k)*4^(k)+4^(k)*4^(k)+4^(k)*4^(k)+4^(k)*4^(k)))
Simplify the FOIL expression by multiplying and combining all like terms.
(2(4294967296))
Complete the multiplication of 2 by each term inside the parenthesis.
(8589934592)
Remove the parenthesis around the expression 8589934592.
8589934592
- 5 years ago
Easy use logs, Log base 4k+4K, 4 then that equals log base 4k+4k to the x over 2 is equal to two. You could use base change to simplify it further but I dunno how simplified you want it. Dude you asked to simplify it and I used my knowledge in calculus. This is the calculus method of simplifying exponents. Logs are exponents or at least have similar traits.
- 1 decade ago
2(4^k + 4^k)^2
= 2*[2(4^k)]^2
= 2[4(4^2k)]
= 2[4^(2k + 1)]
= 2[2^(4k + 2)]
= 2^(4k + 3)
I didn't even look at the answer.
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- 1 decade ago
128^k^2
Source(s): 2(4^k+4^k)(4^k+4^k) using distributive property =(8^k+8^k)(4^k+4^k) =32^k^2 + 32^k^2 + 32^k^2 + 32^k^2 =128^k^2 - Anonymous1 decade ago
Probably want to fix the link, it doesn't go anywhere.
- 1 decade ago
2(4^k +4^k)(4^k +4^k)
2(2)(4^k)(2)(4^k)
8(4^k)
Just a wild guess; couldn't hurt :P
----------------------
Oops...
(2)(2)(2)(4^k)(4^k)
8(4^k)^2 ?
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One more edit:
8 ( 4^2k )
-crosses fingers-
