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2 years ago

13

How many pounds can you loose in a week

2 answers:

I am Lyosha [343]2 years ago

8 0

Depends on the workouts and what you eat too.

nasty-shy [4]2 years ago

7 0

Answer:

well you can lose 10 pounds in a week, but thats not body fat, the average person loses 1 -3 pounds a week with a good diet and exercise.

Explanation:

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Which one of the following statements about mass is most correct?

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Un cualidad en mi caracter es la templanza. Me cuido mucho de lo que vaya a comer. La segunda cualidad positiva mental es que lo que me propongo logro. En el terreno de las relaciones tengo dos. Por ejemplo con mis padres soy bueno hijo y con mi amiga siempre dispuesto de ayudar. En mi grupo de escuela me conocen como una persona amable. Finalmente, en la iglesia yo soy el que ayuda a los ancianos.

Explanation:

One quality of my character is temperance. I take good care of what I am going to eat. The second positive mental quality is whatever I set to accomplish, I will accomplish. In the personal relationships arena, I have two. For example, I am a good son with my parents and my friend I help her in whatever she may need. Within my school, people know me as amicable. Finally, in my church, I help the elderly.

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3 years ago

B(n)=2^n A binary code word of length n is a string of 0's and 1's with n digits. For example, 1001 is a binary code word of len

Citrus2011 [14]

Answer:

The additional words is $2^n$

Explanation:

Given

$B(n) = 2^n$

Required

Determine the additional words; i.e. $B(n + 1) - B(n)$

From the given parameters, we have that;

B is a function of n

Such that;

$B(n) = 2^n$

To calculate $B(n+1)$ , we simply substitute n + 1 for n

$B(n) = 2^n$

$B(n + 1) = 2^{n + 1}$

Applying laws of indices

$B(n + 1) = 2^{n} * 2^1$

$B(n + 1) = 2^{n} * 2$

$B(n + 1) = 2(2^{n})$

Calculating Additional Binary Code;

$B(n + 1) - B(n)$

Substitute values for B(n + 1) and B(n)

$B(n + 1) - B(n) = 2(2^n) - 2^n$

Express $2^n$ as $2^ n * 1$

$B(n + 1) - B(n) = 2(2^n) - 2^n * 1$

Express 1 as $2^0$

$B(n + 1) - B(n) = 2(2^n) - 2^n * 2^0$

Factorize

$B(n + 1) - B(n) = 2^n(2 - 2^0)$

$B(n + 1) - B(n) = 2^n(2 - 1)$

$B(n + 1) - B(n) = 2^n(1)$

$B(n + 1) - B(n) = 2^n$

Hence, the additional words is $2^n$

8 0

2 years ago