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Viefleur [7K]
3 years ago
10

Solve the system of equations. y = 7x + 43 y = 2x + 13 a. ( -6, 1) c. ( 6, 1) b. ( 6, -1) d. No solution

Mathematics
1 answer:
xenn [34]3 years ago
8 0
Hello there!

y = 7x + 43
y = 2x + 13

We wanna solve y = 7x + 43 for y in y = 2x + 13

What does that even mean? Well it simply means that we gonna replace y by 7x + 43.

y = 2x + 13

7x + 43 = 2x + 13

Subtract 2x on both sides

7x + 43 - 2x = 2x + 13 - 2x

5x + 43 = 13

Now we can subtract 43 on both sides

5x + 43 - 43 = 13 - 43

5x = -30

Then divide both sides by 5

5x/5 = -30/5

x = -6

Now since we have the value for x, we gonna use that value to find y.

We can do that by substituting -6 for x in y = 7x + 43

But what does that mean? It means that in y = 7x + 43, we gonna replace the x by -6

y = 7x + 43

y  = 7(-6) + 43

y = -42 + 43

y = 1

Thus,

The answer is: x = -6 and y = 1

The correct answer is option A (-6,1).

Let me know if you have questions about the answer. As always, it is my pleasure to help students like you.
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A pharmacist receives a shipment of 22 bottles of a drug and has 3 of the bottles tested. If 5 of the 22 bottles are contaminate
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Answer:

There is a 44.16% probability that exactly 1 of the tested bottles is contaminated.

Step-by-step explanation:

C_{n,x} is the number of different combinatios of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

In this problem, we have that:

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C_{22,3} = \frac{22!}{3!(18)!} = 1540

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It is 1 one 5(contamined) and 2 of 17(non contamined). So:

C_{5,1}*C_{17,2} = 5*17*8 = 680

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The first two terms of a sequence are 5 and 7. Each term after the second is found by taking the average (arithmetic mean) of al
ch4aika [34]

Answer:

6

Step-by-step explanation:

We can prove that every number after the second will be a six by induction.

<em>Base case.</em> Since \frac{5 + 7}{2} = 6, so the third term is a six.

<em>Inductive hypothesis</em>. Fix the number of terms to be <em>n</em> and assume that

\frac{1}{n} \sum\limits_{i=1}^{n}t_i = 6

<em>Inductive step</em>. We will now show that \frac{1}{n+1} \sum\limits_{i=1}^{n+1}t_i = 6.

Notice that

$\begin{array}{lll}\frac{1}{n+1} \sum\limits_{i=1}^{n+1}t_i & = \frac{n}{n(n+1)} \sum\limits_{i=1}^{n+1}t_i & \\& = \frac{t_{n+1}}{n+1} + \frac{n}{n(n+1)} \sum\limits_{i=1}^{n}t_i &\\& = \frac{t_{n+1}}{n+1} + \frac{6n}{n+1} & \text{(by the IH)}\\& = \frac{6}{n+1} + \frac{6n}{n+1} & \text{by de\\finition}\\& = \frac{6(n+1)}{n+1} & \\& = 6 & \end{array} \square$

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3 years ago
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