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mariarad [96]
2 years ago
15

3^-6x(3^3 divide 3^0)^2

Mathematics
1 answer:
Greeley [361]2 years ago
8 0

Answer:

{3}^{2}

Step-by-step explanation:

3^{ - 6}  \times (3^4  \div  3^0)^2

\frac{1}{  {3}^{6} }  \times ( {3}^{4}  \div  {3}^{0}  {)}^{2}

\frac{1}{729}  \times ( {3}^{4}  \div  {3}^{0}  {)}^{2}

\frac{1}{729}  \times ( {3}^{4}  {)}^{2}

\frac{1}{729}  \times 6561

\frac{1}{729}  \times (729(9))

9

{3}^{2}

<h3>Hope it is helpful...</h3>
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Mohamed decided to track the number of leaves on the tree in his back yard each year. This year there were 500 leaves. Each year
makkiz [27]

Answer:

Geometric sequence

Step-by-step explanation:

According to the scenario, computation of the given data are as follows:

Leaves on tree this year (a) = 500

Rate of increase in leaves per year (r) = 40 % or 0.4

Let n be the number of years.

So we use exponential growth formula, i.e.

f(n) = a(1+r)^n

by putting the value, we get

f(n) = 500 ( 1+0.4)^n

or f(n) = 500 (1.4)^n

So, the common ration is 1.4.

Thus, it shows a geometric sequence.

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3 years ago
Write the product using exponents.<br><br> 2⋅2⋅2⋅n⋅n<br> Using exponents, the product is ?
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The answer is 8n^2

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2 years ago
PLEASE HELP ASAP!!!!!!
zhannawk [14.2K]
The answer is A  to change an exponent from negative to positive, just move it into either the numerator or denominator depending on where it is to begin with. 
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2 years ago
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Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-ma
pickupchik [31]

Answer:

Answer explained below

Step-by-step explanation:

Spam e-mail containing a virus is sent to 1000 e-mail addresses. After 1 second, a recipient machine broadcasts 10 new spam e-mails containing the same virus, after which the virus disables itself on that machine. (1) Write a recursive definition (i.e. recurrence relation) to show how many spam emails will be sent out after n seconds. (2) Solve the recurrence relation. (3) How many e-mails are sent at the end of 20 seconds

1.START T=0.....

1000 EMAILS SENT AND RECEIVED BY 1000 M/CS.

T=1.....

EACH OF THE M/C SENDS 10 NEW MAILS ....

.................................

LET M[N] BE THE NUMBER OF MAILS SENT OUT AFTER N SECONDS.

SO , EACH OF THESE M/CS WILL SEND 10 MAILS IN NEXT 1 SECOND.

HENCE NUMBER OF MAILS SENT IN N+1 SECONDS=M[N+1]=

M[N+1]=10*M[N].......................1

THIS IS THE RECURRENCE RELATION.....

2.SOLUTION .....

M[N+1]=10M[N]=10*10M[N-1]=10*10*10M[N-2]=........

M(N+1)=[10^1][M(N)]=[10^2][M(N-1)]=[10^3][M(N-2)]=..........=[10^N][M(1)]=[10^(N+1)][M(0)]

M[N+1]=[10^(N+1)][1000]=[10^(N+1)][10^3]=[10^(N+4)]......................................2

THIS IS THE NUMBER OF MAILS SENT AFTER N+1 SECONDS .....OR ....

M[N]=[10^(N+3)].............................................3

..................IS THE SOLUTION FOR NUMBER OF MAILS SENT AFTER N SECONDS.....

3.AFTER N=20 SECONDS , THE ANSWER IS ....

M[20]=10^(20+3)=10^23

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Answer: 84,300,000
Simply move the decimal place to the right 7 places
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