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3 years ago

I need help on which ever one you would like to do :)

Mathematics

1 answer:

alexgriva [62]3 years ago

3 0

Answer:

a. k = n - 1 b. 15 c. 112

Step-by-step explanation:

I can answer a. b. and c. for you!

For a. let's say that the variable k means keys. And n means bits.

Since Millie can collect one less keys than there are bits, the equation is....

*See top*

For b. if Antonio has a 16-bit system, by looking at our equation we know that k = n - 1.

So, if n = 16, k = 15

For c. if each byte equals 8 bits, and there are 14 bytes, then the amount of bits would be 8 x 14.

8 x 14 = 112

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Alex_Xolod [135]

The answer is A) 6:15, because if you divide both numbers by three, it'll give you 2:5

4 0

3 years ago

Read 2 more answers

Which expression is equivalent to −19y+13 ?

Vikki [24]

The first thing you should do in this case is to write the expression correctly:
- (1/9) y + (1/3)
Now, we take out common factor (1/9) obtaining:
1/9 (-y + 3)
Let's check:
1/9 (-y + 3)
-1/9y + 3/9
-1/9y + 1/3
OK
Answer:
An expression equivalent is:
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4 0

2 years ago

Read 2 more answers

PLEASE RESPOND! !!

cestrela7 [59]

Answer:

36/132

Step-by-step explanation:

36/132, .27, 27%...

6 0

3 years ago

The fitness club opens with 80 members each month the membership Increases by 15 members which equation represents the relations

creativ13 [48]

Answer:

y=15x+80

Step-by-step explanation:

where x is the monthly membership increase and y is the total amount of memberships

3 0

3 years ago

Ask your teacher find the limit. use l'hospital's rule where appropriate. if there is a more elementary method, consider using i

liberstina [14]

Answer:

ln(5/3)

Step-by-step explanation:

The desired limit represents the logarithm of an indeterminate form, so L'Hopital's rule could be applied. However, the logarithm can be simplified to a form that is not indeterminate.

<h3>Limit</h3>

We can cancel factors of (x-1), which are what make the expression indeterminate at x=1. Then the limit can be evaluated directly by substituting x=1.

$\diplaystyle \lim\limits_{x\to1}{(\ln(x^5-1)-\ln(x^3-1))}=\lim\limits_{x\to1}\ln{\left(\dfrac{x^5-1}{x^3-1}\right)}\\\\=\lim\limits_{x\to1}\ln\left(\dfrac{x^4+x^3+x^2+x+1}{x^2+x+1}\right)=\ln{\dfrac{5}{3}}$

8 0

1 year ago