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guajiro [1.7K]
3 years ago
9

Which quadratic equation is equivalent to (x-2)^2+5(x+2)-6=0

Mathematics
2 answers:
Juliette [100K]3 years ago
5 0

Answer:

x² + x + 8 = 0

Step-by-step explanation:

(x-2)^2 + <u>5(x+2) </u>- 6=0                     { (a-b)² = a² -2ab + b²; here a = x & b = 2}

x²- 2*x*2 + 2² + <u>5 x + 10</u> -6 = 0

x² - 4x +<u> </u>4 + 5x + 10 - 6 = 0

x² + x + 4 + 10 - 6 = 0

x² + x + 8 = 0

Setler [38]3 years ago
4 0

Answer:

x² + x + 8 = 0

Step-by-step explanation:

Given

(x - 2)² + 5(x + 2) - 6 = 0 ← expand (x - 2)² and distribute parenthesis by 5

x² - 4x + 4 + 5x + 10 - 6 = 0 ← collect like terms on left side

x² + x + 8 = 0 ← equivalent quadratic equation

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Answer:

x = -3

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Step-by-step explanation:

y = -2x + 1

7 = -2x + 1

<u>-1           -1 </u>

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divide by -2

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Find the solution to the initial value problem
Karo-lina-s [1.5K]

Answer:

y = (11x + 13)e^(-4x-4)

Step-by-step explanation:

Given y'' + 8y' + 16 = 0

The auxiliary equation to the differential equation is:

m² + 8m + 16 = 0

Factorizing this, we have

(m + 4)² = 0

m = -4 twice

The complimentary solution is

y_c = (C1 + C2x)e^(-4x)

Using the initial conditions

y(-1) = 2

2 = (C1 -C2) e^4

C1 - C2 = 2e^(-4).................................(1)

y'(-1) = 3

y'_c = -4(C1 + C2x)e^(-4x) + C2e^(-4x)

3 = -4(C1 - C2)e^4 + C2e^4

-4C1 + 5C2 = 3e^(-4)..............................(2)

Solving (1) and (2) simultaneously, we have

From (1)

C1 = 2e^(-4) + C2

Using this in (2)

-4[2e^(-4) + C2] + 5C2 = 3e^(-4)

C2 = 11e^(-4)

C1 = 2e^(-4) + 11e^(-4)

= 13e^(-4)

The general solution is now

y = [13e^(-4) + 11xe^(-4)]e^(-4x)

= (11x + 13)e^(-4x-4)

3 0
3 years ago
Triangle ABC has angle measures as shown below. Using the information in the diagram, and the value of x that you calculated in
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Answer:

x=29

Step-by-step explanation:

180-87=93

93=3x+6

-6        -6

--------------

93=3x

/3    /3

-------------

29=x

x=29

*all angles in a triangle add up to 180. So, 35+52=93. 180-93=87.

C=87

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