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Lesechka [4]
3 years ago
8

What property is shown in the equation 7x divided by 1 = 7x

Mathematics
1 answer:
victus00 [196]3 years ago
6 0

Answer:

all real numbers

Step-by-step explanation:


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Solve 8[7-(6x-6)]+6x=0.
zavuch27 [327]
I believe that X would be -2.47619.
 First you distribute the -1 in front of the parenthesis. Then distribute the 8 outside of the bracts. And finally just solve for X by combing like terms.
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3 years ago
Evaluate the line integral, where C is the given curve. (x + 6y) dx + x2 dy, C C consists of line segments from (0, 0) to (6, 1)
Dima020 [189]

Split C into two component segments, C_1 and C_2, parameterized by

\mathbf r_1(t)=(1-t)(0,0)+t(6,1)=(6t,t)

\mathbf r_2(t)=(1-t)(6,1)+t(7,0)=(6+t,1-t)

respectively, with 0\le t\le1, where \mathbf r_i(t)=(x(t),y(t)).

We have

\mathrm d\mathbf r_1=(6,1)\,\mathrm dt

\mathrm d\mathbf r_2=(1,-1)\,\mathrm dt

where \mathrm d\mathbf r_i=\left(\dfrac{\mathrm dx}{\mathrm dt},\dfrac{\mathrm dy}{\mathrm dt}\right)\,\mathrm dt

so the line integral becomes

\displaystyle\int_C(x+6y)\,\mathrm dx+x^2\,\mathrm dy=\left\{\int_{C_1}+\int_{C_2}\right\}(x+6y,x^2)\cdot(\mathrm dx,\mathrm dy)

=\displaystyle\int_0^1(6t+6t,(6t)^2)\cdot(6,1)\,\mathrm dt+\int_0^1((6+t)+6(1-t),(6+t)^2)\cdot(1,-1)\,\mathrm dt

=\displaystyle\int_0^1(35t^2+55t-24)\,\mathrm dt=\frac{91}6

6 0
3 years ago
PLZ HELP ASAP!!!!!!!!!!!
lorasvet [3.4K]
For number 3 add all the numbers and divide how numbers you have
5 0
3 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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3 years ago
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Simple use the cuboid volume formula for each of them
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