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Natasha_Volkova [10]
3 years ago
8

Which do you pick and why? How much money do you end up with in each case?

Mathematics
1 answer:
Mademuasel [1]3 years ago
7 0

Answer:

1. Start with $1 and then double the money you have everyday for 30 days. You would end up with 1,073,741,824 on the 30th day.

Step-by-step explanation:

Why you should choose the first option:

1 x 2 = 2

2 x 2 = 4

4 x 2 = 8

8 x 2 = 16

16 x 2 = 32

32 x 2 = 64

64 x 2 = 128

128 x 2 = 256

256 x 2 = 512

512 x 2 = 1024

1024 x 2 = 2048

2048 x 2 = 4096

4096 x 2 = 8192

8192 x 2 = 16384

16384 x 2 = 32768

32768 x 2 = 65536

65536 x 2 = 131072

131072 x 2 = 262144

262144 x 2 = 524288

524288 x 2 = 1048576

1048576 x 2 = 2097152

2097152 x 2 = 4194304

4194304 x 2 = 8388608

8388608 x 2 = 16777216

16777216 x 2 = 33554432

33554432 x 2 = 67108864

67108864 x 2 = 134217728

134217728 x 2 = 268435456

268435456 x 2 = 536870912

536870912 x 2 = 1073741824

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a. Use the graphical or eigenvalue approach

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What is the least common denominator for 5,6,and7?
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Answer:

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Step-by-step explanation:

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7 0
3 years ago
X - y = 2<br> 4x – 3y = 11
swat32

Answer:

<u><em>For x - y = 2</em></u>

You need to find x or y in one of the equations and then substitute that into the other.

So we have;

x-y=2

4x-3y=11

We will take the first equation and find x;

x-y=2

add y to both sides;

x-y+y=2+y

x=2+y

Now we take that answer and substitute it forx in the other equation;

 

4(2+y)-3y=11

8+4y-3y=11

8+y=11

y=3

Now we have what y equals, so we use it in the first equation to find x;

x-3=2

x=5

So we have;

x=5; y=3

Hope you understand!

=)

<u><em>And for 4x – 3y = 11</em></u>

Multiply the first equation by 2 and the second by 3 so that there are the same number of y's in each:

8x - 6y = 22    ...(3)

30x + 6y = -3  ...(4)

 Now add (3) and (4) term by term:

38x + 0 = 19

or

38x = 19

or x = 1/2

Put this back into equation (1)

4*(1/2) - 3y = 11

or

2 - 3y = 11

Subtract 2 from both sides:

-3y = 9

 Divide both sides by -3

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6 0
3 years ago
Read 2 more answers
Determine determine whether the following geometric series converges or diverges. if the series converges find its sum.
Lilit [14]

For starters,

\dfrac{3^k}{4^{k+2}}=\dfrac{3^k}{4^24^k}=\dfrac1{16}\left(\dfrac34\right)^k

Consider the nth partial sum, denoted by S_n:

S_n=\dfrac1{16}\left(\dfrac34\right)+\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\cdots+\dfrac1{16}\left(\dfrac34\right)^n

Multiply both sides by \frac34:

\dfrac34S_n=\dfrac1{16}\left(\dfrac34\right)^2+\dfrac1{16}\left(\dfrac34\right)^3+\dfrac1{16}\left(\dfrac34\right)^4+\cdots+\dfrac1{16}\left(\dfrac34\right)^{n+1}

Subtract S_n from this:

\dfrac34S_n-S_n=\dfrac1{16}\left(\dfrac34\right)^{n+1}-\dfrac1{16}\left(\dfrac34\right)

Solve for S_n:

-\dfrac14S_n=\dfrac3{64}\left(\left(\dfrac34\right)^n-1\right)

S_n=\dfrac3{16}\left(1-\left(\dfrac34\right)^n\right)

Now as n\to\infty, the exponential term will converge to 0, since r^n\to0 if 0. This leaves us with

\displaystyle\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=1}^n\frac{3^k}{4^{k+2}}=\sum_{k=1}^\infty\frac{3^k}{4^{k+2}}=\frac3{16}

8 0
3 years ago
Read 2 more answers
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