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gregori [183]
3 years ago
5

8x-5=-3(3-2x) Sole for x

Mathematics
2 answers:
mario62 [17]3 years ago
6 0
Solve for x:  <span>8x-5=-3(3-2x)

1.  Carry out the indicated multiplication:  8x - 5 = - 9 + 6x
2.  Combine like terms:  2x = -4
3.  Solve this final equation for x:     x = -2         (answer)

</span>
horrorfan [7]3 years ago
4 0
The first step would be to simplify both sides of the equal sign as much as possible. 
8x - 5 = -3(3 - 2x)
8x - 5 = -9 - (-6x)
8x - 5 = -9 + 6x

Next, you would need to combine like terms on both sides of the equal sign to one on one side of the equal sign and one on the other. 
8x - 5 = -9 + 6x
2x - 5 = -9
2x = -4

Now, all you have to do is isolate the x. To do this, you would divide both sides by 2. 
2x = -4
x = -2

I hope this helps!

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There are 10 cards. Each card has one number between 1 and 10, so that every number from 1 to 10 appears once.
pychu [463]

Answer:

When a card is chosen at random with replacement five times, X is the number of times a prime number is chosen.          Here the card is chosen with replacement.  This implies probability for choosing a prime number remains the same as the previously drawn card is replaced.

The sample space= {1,2,3,4,5,6,7,8,9,10}

Prime numbers = {2,3,5,7}

Prob for drawing prime number = 4/10 = 0.4

is the same when replacement is done.

Also there are two outcomes either prime or non prime.  Hence in this case, X the no of times a prime number is chosen, is binomial with p =0.4 and q = 0.6 and n=5


When a card is chosen at random without replacement three times, X is the number of times an even number is chosen.

Prob for an even number = 0.5

But after one card drawn say odd number next card has prob for even number as 5/9 hence each draw is not independent of the other.  Hence not binomial.

When a card is chosen at random with replacement six times, X is the number of times a 3 is chosen.

Here since every time replacement is done, probability of drawing a 3 remains constant = 1/10 = 0.3

i.e. each draw is independent of the other and there are only two outcomes , 3 or non 3. Hence here X is binomial.

When a card is chosen at random with replacement multiple times, X is the number of times a card is chosen until a 5 is chosen

Here X is the number of times a card is chosen with replacement till 5 is chosen.  This is not binomial.  Here probability for drawing nth time correct 5 is  P(non 5 in the first n-1 draws)*P(5 in nth draw) = 0.1^(n-1) (0.9)

Because nCr is not appearing i.e. 5 cannot appear in any order but only in the last draw, this is not binomial.

Step-by-step explanation:


6 0
2 years ago
Sequence 1,2,3,6,18,108 multiply the previous 2 terms to get the next term
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2 years ago
Suppose you solved a second-order equation by rewriting it as a system and found two scalar solutions: y = e^5x and z = e^2x. Th
xenn [34]

Answer:

The solutions are linearly independent because the Wronskian is not equal to 0 for all x.

The value of the Wronskian is \bold{W=-3e^{7x}}

Step-by-step explanation:

We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.

W = \left|\begin{array}{cc}y&z\\y'&z'\end{array}\right|

Thus replacing the functions of the exercise we get:

W = \left|\begin{array}{cc}e^{5x}&e^{2x}\\5e^{5x}&2e^{2x}\end{array}\right|

Working with the determinant we get

W = 2e^{7x}-5e^{7x}\\W=-3e^{7x}

Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.

3 0
3 years ago
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