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Setler79 [48]
3 years ago
12

Find the solutions of the quadratic equation x2 + 7x + 10 = 0.

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
7 0

Answer:

Step-by-step explanation:

x² + 7x + 10 = 0

x = [-7 ± √(7² - 4·1·10)]/(2·1) = [-7 ± √9[/2 = [-7 ± 3]/2 = -2, -5

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tekilochka [14]
The film lasted until 8:27.

This is true because 102 minutes is 1 hour and 42 minutes. If you add 1 hour and 42 minutes onto 6 hours and 45 minutes, the beginning time then. You would get 8 hours and 27 minutes.

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   7:87 <---- But that isn't correct because 87 minutes is over an hour.  

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5 0
3 years ago
Give a geometric description of the following system of equations.a. 2x−4y=12 −3x+6y=−15.b. 2x−4y=12 −5x+3y=10.a. 2x−4y=12 −3x+6
Reptile [31]

Answer:

a. No solution, parallel lines.

b. One solution.

Step-by-step explanation:

Given the system of equations:

a. 2x-4y=12

-3x+6y=-15

b. 2x-4y=12

-5x+3y=10

To give a geometric description of the given system of equations.

The geometric description of a system of equations in 2 variables mean the system of equations will represent the number of lines equal to the number of equations in the system given.

i.e.

Number of planes = Number of variables

Number of lines = Number of equations in the system.

Here, we are given 2 variables and 2 equation in each system.

So, they can be represented in the xy-coordinates plane.

And the number of solutions to the system depends on the following condition.

Let the system of equations be:

A_1x+B_1y+C_1=0\\A_2x+B_2y+C_2=0

1. One solution:

There will be one solution to the system of equations,  If we have:

\dfrac{A_1}{A_2}\neq\dfrac{B_1}{B_2}

2. Infinitely Many Solutions: (Identical lines in the system)

\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}= \dfrac{C_1}{C_2}

3. No Solution:(Parallel lines)

\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}\neq\dfrac{C_1}{C_2}

Now, let us discuss the system of equations one by one:

a. 2x-4y=12 OR 2x-4y-12=0

-3x+6y=-15 OR -3x+6y+15=0

A_1 = 2, B_1 = -4, C_1 = -12\\A_2 = -3, B_2 = 6, C_2= 15

Here, the ratio:

\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2} = -\dfrac{2}{3}\\\dfrac{C_1}{C_2} = -\dfrac{4}{5}

\dfrac{A_1}{A_2}=\dfrac{B_1}{B_2}\neq\dfrac{C_1}{C_2}

Therefore, no solution i.e. parallel lines.

b. 2x-4y=12 OR 2x-4y-12=0

-5x+3y=10 OR -5x+3y-10=0

A_1 = 2, B_1 = -4, C_1 = -12\\A_2 = -5, B_2 = 3, C_2 = -10

\dfrac{A_1}{A_2}= -\dfrac{2}{5}\\\dfrac{B_1}{B_2} = -\dfrac{4}{3}\\\dfrac{C_1}{C_2} = -\dfrac{6}{5}

\dfrac{A_1}{A_2}\neq\dfrac{B_1}{B_2}

So, one solution.

Kindly refer to the images attached for the graphical representation of the given system of equations.

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