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(2^(4-1)^2)/(2^(3-2))
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(2^(4-1)^2)/(2^(3-2))
Fri Nov 10, 2017 6:12 am
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Fri Nov 10, 2017 7:28 am
Hello Vincen.
Let's do the calculus step by step.
$$1-\,\, \left(4-1\right)^2=\left(3\right)^2=9.$$ So, $$2^{\left(4-2\right)^2}=2^9.$$ Now, $$2-\,\, \ 2^{\left(3-2\right)}=2^1.$$ Finally, $$3-\,\, \ \frac{2^{\left(4-1\right)^2}}{2^{\left(3-2\right)}}=\frac{2^9}{2^1}=2^{\left(9-1\right)}=2^{8.}$$ So, the correct answer is A .
I hope this can help you.
I'm available if you'd like a follow up.
Let's do the calculus step by step.
$$1-\,\, \left(4-1\right)^2=\left(3\right)^2=9.$$ So, $$2^{\left(4-2\right)^2}=2^9.$$ Now, $$2-\,\, \ 2^{\left(3-2\right)}=2^1.$$ Finally, $$3-\,\, \ \frac{2^{\left(4-1\right)^2}}{2^{\left(3-2\right)}}=\frac{2^9}{2^1}=2^{\left(9-1\right)}=2^{8.}$$ So, the correct answer is A .
I hope this can help you.
I'm available if you'd like a follow up.
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Fri Nov 10, 2017 8:30 am
Vincen wrote:
I have a feeling that you didn't mean to write that the expression $$\frac{2^{\left(4-1\right)^2}}{2^{\left(3-2\right)}}$$ equals ZERO.
$$\frac{2^{\left(4-1\right)^2}}{2^{\left(3-2\right)}}=0$$
A. 2^8
B. 2^7
C. 2^6
D. 2^5
E. 2^4
The OA is A .
I don't know how to do the calculus here. Experts, can you help me?
Thanks.
A. 2^8
B. 2^7
C. 2^6
D. 2^5
E. 2^4
The OA is A .
I don't know how to do the calculus here. Experts, can you help me?
Thanks.
This would be impossible.
Cheers,
Brent
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