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bazaltina [42]
3 years ago
15

Solve the following system of equations algebraically: y= x² + 11x + 6 y = 2x – 8

Mathematics
1 answer:
EleoNora [17]3 years ago
7 0
4 I think it’s 4 because 2x would be 4 and 8
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1/2 of n decreased by 8
Sergeu [11.5K]

Answer:

1/2n - 8

Step-by-step explanation:

1/2 of n  means 1/2n

decreased by 8 means subtract 8

1/2n - 8

7 0
3 years ago
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Question 1 of 10
AleksAgata [21]

Answer:

True

Step-by-step explanation:

3 0
3 years ago
Question is attached
rodikova [14]

9514 1404 393

Answer:

  x = 12

  perimeter = 124

Step-by-step explanation:

The midline RS is half the length of MN, so we have ...

  2×RS = MN

  2(x +3) = 5x -30

  2x +6 = 5x -30

  36 = 3x . . . . . . . . . add 30-2x

  12 = x . . . . . . . . . . divide by 3

__

The length RS is then ...

  RS = x +3 = 15

and the perimeter of QRS is ...

  P(QRS) = QR +RS +SQ = 25 +15 +22 = 62

The perimeter of QRS is half the perimeter of MNP, so ...

  P(MNP) = 2×P(QRS) = 2×62 = 124

The perimeter of ΔMNP = 124.

8 0
2 years ago
The sum of three consecutive numbers is 84. What is the largest of these numbers?
just olya [345]
Let the numbers = x, x+1, x+2
It is given that, x + x+1 + x+2 = 84
3x + 3 = 84
3x = 84 - 3
3x = 81
x = 81/3
x = 27
Largest number, x+2 = 27+2 = 29

In short, Your Answer would be 29

Hope this helps!
6 0
3 years ago
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Consider the differential equation y '' − 2y ' + 37y = 0; ex cos(6x), ex sin(6x), (−[infinity], [infinity]). Verify that the giv
givi [52]

The question  is:

Consider the differential equation y'' - 2y' + 37y = 0;

e^x cos(6x), e^x sin(6x), (-\infty,  \infty).

Verify that the given functions form a fundamental set of solutions of the differential equation on the indicated interval.

The functions satisfy the differential equation and are linearly independent since

a) W(e^x cos 6x, e^x sin 6x) \neq 0

Answer:

To verify if the given functions form a fundamental set of solutions to the differential equation, we find the Wronskian of the two functions.

The Wronskian of functions y_1 $and$ y_2 is given asW(y_1, y_2) = \left|\begin{array}{cc}y_1&y_2\\y_1'&y_2'\end{array}\right|\\\\y_1= e^x cos 6x \\y_2 = e^x sin 6x \\y_1' = -6e^x sin 6x \\y_2'  = 6e^x cos 6x \\W\left(e^x cos 6x, e^x sin 6x \right) =  \left|\begin{array}{cc}e^x cos 6x &e^x sin 6x \\ \\ -6e^x sin 6x&6e^x cos 6x \end{array}\right|\\ \\=  6e^{2x} cos^2 6x + e^{2x} sin^2 6x \\ \\ = 6e^{2x}\left( cos^2 6x + sin^2 6x\right)\\ \\$but $cos^2 6x + sin^2 6x = 1\\ \\W\left(y_1, y_2 \right) = 6e^{2x} \neq 0

Because the Wronskian is not zero, we say the solutions are linearly independent

3 0
4 years ago
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