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2 years ago

15 _ { 10 } + X _ { 2 } = 11011 _ { 2 }

Mathematics

1 answer:

KatRina [158]2 years ago

5 0

11011₂ = 2⁴ + 2³ + 2¹ + 2⁰ = 27

15 + x₂ = 27

x₂ = 12 = 8 + 4 = 2³ + 2² = 1100₂

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Complete the table below by calculating the commission and adding it to the price of the stock.

asambeis [7]

We calculate the commission and adding it to the price of the stock to get total cost.

No Price Total Commission Total cost

100 $26.25 0.06 x $2,625 = $157.50 2625 + 157.50 =2782.5

100 $19.32 0.06 x $1,932 = $115.92 1932 + 115.92 =2047.92

40 $9.77 0.06 x $390.80 = $23.45 390.80 + 23.45= 414.25

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100 $33.44 0.06 x $3,344 = $200.64 3,344 + 200.64 = 3544.64

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3 years ago

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Fin the values of x and y, help me

astraxan [27]

Answer:

x = 25 , y = 19

Step-by-step explanation:

Since the triangle has 3 congruent sides then it is equilateral

The 3 angles are also congruent, with each angle = 60° , then

2x + 10 = 60 ( subtract 10 from both sides )

2x = 50 ( divide both sides by 2 )

x = 25

and

3y + 3 = 60 ( subtract 3 from both sides )

3y = 57 ( divide both sides by 3 )

y = 19

3 0

3 years ago

A bicycle is originally priced at $80 the store owner gives a discount in the bicycle is know priced at $60 enter the percent ma

Roman55 [17]

Answer:

25%

Step-by-step explanation:

Since we know that the markdown is $20 by subtracting 80-60 , we find out what percent of 20 is 80. To do that you do 20/80, which is 0.25, or 25 percent.

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3 years ago

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Solve the equation. 5x + 8 − 3x = −10

inna [77]

Answer:x=-9

Step-by-step explanation:

7 0

3 years ago

Lim (n/3n-1)^(n-1) n → ∞

n200080 [17]

Looks like the given limit is

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}$

With some simple algebra, we can rewrite

$\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)$

then distribute the limit over the product,

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}$

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, e :

$\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n$

To make our limit resemble this one more closely, make a substitution; replace 9/(n - 9) with 1/m, so that

$\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1$

From the relation 9m = n - 9, we see that m also approaches infinity as n approaches infinity. So, the second limit is rewritten as

$\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}$

Now we apply some more properties of multiplication and limits:

$\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9$

So, the overall limit is indeed 0:

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}$

7 0

3 years ago

15 _ { 10 } + X _ { 2 } = 11011 _ { 2 }​

15 _ { 10 } + X _ { 2 } = 11011 _ { 2 }