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MatroZZZ [7]
3 years ago
14

Finish this 1-5 all of it

Mathematics
1 answer:
horrorfan [7]3 years ago
3 0
You are lazy... do it on your own.
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What is the sum of the answers to the eqation?<br><img src="https://tex.z-dn.net/?f=%20%7Bx%7D%5E%7B2%7D%20%20-%206x%20%2B%205%2
MAVERICK [17]

Answer:

6

Step-by-step explanation:

x^{2} - 6x +5 = 0

x^{2} - 6x + 9 -9 +5 = 0

[x^{2} - 6x + 9] - 4 =0

(x -3)^{2} -4 = 0

(x -3)^{2} = 4

(x-3) = \sqrt{4} or (x-3) = - \sqrt{4}

(x-3) = 2 or (x-3) = -2

x = 5 or x = 1

therefore, sum of these solutions is 5+1 = 6

4 0
3 years ago
Read 2 more answers
Z = 18ba - 6b<br><br> Solve for a
adelina 88 [10]

Answer:

   

Step-by-step explanation:

       

7 0
3 years ago
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PLEASE ANSWER AS SOON AS YOU SEE THIS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Which expression is equivalent to 5 (2 + 7)? 2 (5 + 7) 2 + 7
mamaluj [8]

Answer:

  • D. 5 (2) + 5 (7)

Step-by-step explanation:

<u>Given expression:</u>

  • 5(2 + 7)

<u>Distribute this to get:</u>

  • a(b + c) = ab + ac
  • 5(2 + 7) = 5(2) + 5(7)

Correct choice is D

6 0
3 years ago
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Help me please <br> And good morning
konstantin123 [22]

Answer:

It should be -8

Step-by-step explanation:

5 0
3 years ago
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Determine the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solutio
AnnZ [28]

Answer:

y'' + \frac{7}{t} y = 1

For this case we can use the theorem of Existence and uniqueness that says:

Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:

y''+ p(t) y' +q(t) y = g(t) , y(t_o) =y_o, y'(t_o) = y'_o

has unique solution defined for all t in [a,b]

If we apply this to our equation we have that p(t) =0 and q(t) = \frac{7}{t} and g(t) =1

We see that q(t) is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by (0, \infty)

And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.

Step-by-step explanation:

For this case we have the following differential equation given:

t y'' + 7y = t

With the conditions y(1)= 1 and y'(1) = 7

The frist step on this case is divide both sides of the differential equation by t and we got:

y'' + \frac{7}{t} y = 1

For this case we can use the theorem of Existence and uniqueness that says:

Let p(t) , q(t) and g(t) be continuous on [a,b] then the differential equation given by:

y''+ p(t) y' +q(t) y = g(t) , y(t_o) =y_o, y'(t_o) = y'_o

has unique solution defined for all t in [a,b]

If we apply this to our equation we have that p(t) =0 and q(t) = \frac{7}{t} and g(t) =1

We see that q(t) is not defined at t =0, so the largest interval containing 1 on which p,q and g are defined and continuous is given by (0, \infty)

And by the theorem explained before we ensure the existence and uniqueness on this interval of a solution (unique) who satisfy the conditions required.

7 0
3 years ago
Read 2 more answers
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