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tankabanditka [31]
3 years ago
9

0=x^2-4x+3 how many roots does this equation have

Mathematics
1 answer:
mars1129 [50]3 years ago
5 0

Answer:

No complex roots; 3 real roots

Step-by-step explanation:

If a third order polynomial has any complex roots, then as a rule it has 1 real root and 2 complex roots.  In this particular case, the polynomial has three real roots, as can be determined by graphing the function.  The graph crosses the x-axis in 3 places.

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ratelena [41]

Answer:

x=1/6

Step-by-step explanation:

1/2-1/3=x

x=1×3-1×3/6 [ take l.c.m]

x=3-2/6

x=1/6 .

8 0
2 years ago
Please help !!<br> 12x + 14 = 6x + 7 (+) 9x - 5 find the value of x
Aloiza [94]

Answer:

x=4

Step-by-step explanation:

6 0
3 years ago
What is two and two fourths plus one third
kkurt [141]

Answer:

2 5/6

Step-by-step explanation:

2 2/4 + 1/3

First, you have to make the bottoms of the fraction the same, by figuring out the lowest common denominator. In this case it would be 12. 4x3 = 12. 3x4 = 12.

Multiply the top number by the same number you multiplied the bottom by.

We multiplied the 4 by 3, so we would also multiply the 2 by 3 (which would be 6).

Do the same for the second fraction. 1x4 = 4.

Now we have 2 6/12 + 4/12. We add the top numbers together and we get 10/12.

Now we have to reduce the fractions. We can do this in this situation by just dividing the top and bottom numbers by 2.

2 5/6

8 0
3 years ago
Y=-x^2-10x-16 on graph
lana [24]

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It is a Non-Linear graph:

8 0
3 years ago
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

8 0
3 years ago
Read 2 more answers
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