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kotykmax [81]
3 years ago
10

What are the solutions to the system of equations? {y=x²−4x+8 y=2x+3

Mathematics
1 answer:
MrRa [10]3 years ago
6 0

Answer:

x = 1, y = 5

x = 5, y = 13

Step-by-step explanation:

y = x² − 4x + 8 .......(1)

y = 2x + 3 .......(2)

Substitute the value of y in equation 2 into equation 1

y = x² − 4x + 8

y = 2x + 3

x² − 4x + 8 = 2x + 3

Rearrange

x² − 4x − 2x + 8 − 3 = 0

x² − 6x + 5 = 0

Solve by factorization

Find the product you x² and 5. The result is 5x²

Find the factors of 5x² such that their sum will result in −6x. The factors are −x and −5x.

Replace −6x in the equation above with −x and −5x. This is illustrated below:

x² − x − 5x + 5 = 0

x(x − 1) − 5(x − 1) = 0

(x − 1)(x − 5) = 0

x − 1 = 0 or x − 5 = 0

x = 1 or x = 5

Substitute the value of x into equation 2 to obtain the value of y.

y = 2x + 3

x = 1

y = 2(1) + 3

y = 2 + 3

y = 5

x = 5

y = 2x + 3

y = 2(5) + 3

y = 10 + 3

y = 13

SUMMARY:

x = 1, y = 5

x = 5, y = 13

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Use the half life information to find k:

\dfrac{1}{2}P_o=P_oe^{k(800)}\\\\\\\dfrac{1}{2}=e^{800k}\qquad \rightarrow \qquad \text{divided both sides by}\ P_o\\\\\\ln\bigg(\dfrac{1}{2}\bigg)=800k\qquad \rightarrow \qquad \text{applied ln to both sides}\\\\\\\dfrac{ln(\frac{1}{2})}{800}=k\qquad \rightarrow \qquad \text{divided both sides by 800}\\\\\\\large\boxed{-0.000867=k}

Next, input the information (65% decayed = 35% remaining) and the k-value to find your answer.

.35P_o=P_oe^{-0.000867t}\\\\\\.35=e^{-0.000867t}\\\\\\ln(.35)=-0.000867t\\\\\\\dfrac{ln(.35)}{-0.000867}=t\\\\\\\large\boxed{1211.6585=t}

5 0
3 years ago
An earthquake with a rating of 3.3 is not
stiks02 [169]

Answer:

  1995.26

Step-by-step explanation:

It usually works to follow instructions:

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This is Matrix for pre calc
gavmur [86]

The given system of equations in augmented matrix form is

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\-6&1&2&4&-12\\1&-3&-3&5&-20\\-2&5&6&0&12\end{array}\right]

If you need to solve this, first get the matrix in RREF:

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\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&11&5&-13&37\\0&19&10&4&-10\end{array}\right]

  • Add 11(row 2) to -5(row 3), and 19(row 1) to -5(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&-164&132&-1052\end{array}\right]

  • Add 164(row 3) to -91(row 4):

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&13080&-39240\end{array}\right]

  • Multiply row 4 by 1/13080:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&153&-823\\0&0&0&1&-3\end{array}\right]

  • Add -153(row 4) to row 3:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&-91&0&-364\\0&0&0&1&-3\end{array}\right]

  • Multiply row 3 by -1/91:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&-6&8&-58\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Add 6(row 3) and -8(row 4) to row 2:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&5&0&0&-10\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Multiply row 2 by 1/5:

\left[\begin{array}{cccc|c}3&2&-4&2&-23\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Add -2(row 2), 4(row 3), and -2(row 4) to row 1:

\left[\begin{array}{cccc|c}3&0&0&0&3\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

  • Multiply row 1 by 1/3:

\left[\begin{array}{cccc|c}1&0&0&0&1\\0&1&0&0&-2\\0&0&1&0&4\\0&0&0&1&-3\end{array}\right]

So the solution to this system is (w,x,y,z)=(1,-2,4,-3).

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3 years ago
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spayn [35]
7(x+4)>0

Distribute 7. (multiply by x and 4)

7x+28>0

Subtract 28 on both sides.

7x>-28

Divide by 7 into both sides.

x>-4

x>-4 is the answer. 

I hope this helps! 
~kaikers
7 0
3 years ago
Read 2 more answers
What is the value of the expression |a + b| + |c| when a = –6, b = 2, and c = –11?
nika2105 [10]
|-6 + 2| + |-11| 

|4| + |-11|

= 15

Hope i helped! 
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3 years ago
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