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Y_Kistochka [10]
3 years ago
5

Use the quadratic formula to solve x2 + 9x + 10 = 0. What are the solutions to the equation?

Mathematics
2 answers:
vampirchik [111]3 years ago
5 0

Answer:

\large\boxed{x=\dfrac{-9\pm\sqrt{41}}{2}}

Step-by-step explanation:

\text{The quadratic formula of}

ax^2+bx+c=0

\text{If}\ b^2-4a0,\ \text{then the equation has two solutions}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\=========================================

\text{We have}\ x^2+9x+10=0\\\\a=1,\ b=9,\ c=10\\\\\text{substitute:}\\\\b^2-4ac=9^2-4(1)(10)=81-40=41>0\qquad _{\text{two solutions}}\\\\\sqrt{b^2-4ac}=\sqrt{41}\\\\x=\dfrac{-9\pm\sqrt{41}}{2(1)}=\dfrac{-9\pm\sqrt{41}}{2}

HACTEHA [7]3 years ago
3 0

Answer:

<h3>The solution of given quadratic equation  =  [-9 ± √41]/2</h3>

Step-by-step explanation:

Points to remember

solution of a quadratic equation ax² + bx + c = 0

x = [-b ± √(b² - 4ac)]/2a

It is given that, x² + 9x + 10 = 0.

<u>To find the solution</u>

Here a = 1, b= 9 and c = 10

x = [-b ± √(b² - 4ac)]/2a

 = [-9 ± √(9² - 4*1*10)]/2*1

 = [-9 ± √(81 - 40)]/2

 =  [-9 ± √(41)]/2

 = [-9 ± √41]/2

Therefore the solution of given quadratic equation  =  [-9 ± √41]/2

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Answer:

Part 1:

The solution set of the system of equations is x = 1, y = 1

Part 2:

The solution set is x = 1, y = 1

Part 3:

The similarity of the system of equations in Parts 1 and 2 is that they have the same solution set

The difference of the system of equations in Parts 1 and 2 is that they are arranged differently

Step-by-step explanation:

Part 1:

The system of equation is given as follows;

2·x + y = 3...(1)

x = 2·y - 1...(2)

The above system of equations can be written in terms of the variable, y, as follows;

For equation (1), we have;

y = 3 - 2·x

For equation (2), we have;

y = (x + 1)/2

From the attached graph created with Microsoft Excel, we have;

The solution set (the point of intersection) of the system of equations is x = 1, y = 1

To accurately find the common solution, we have;

(x + 1)/2 = 3 - 2·x

x + 1 = 2·(3 - 2·x) = 6 - 4·x

x + 1 = 6 - 4·x

x + 4·x = 6 - 1

5·x = 5

x = 5/5 = 1

x = 1

Therefore, y = 3 - 2·x = 3 - 2× 1 = 1, at the common solution

Part 2:

y = -2x + 3

x - 2y = -1

∴ x = 2y -1

y = -2(2y -1) + 3

y = -4y + 2 + 3

5y = 5

y = 1

x = 2y - 1 = 2 - 1 = 1

x = 1

The solution set is x = 1, y = 1

Part 3

The system of equations are similar in terms of their solution

The system of equations are different in terms of their arrangement

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Step-by-step explanation:

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What is the height of a cylinder with a volume of 936 times 3.14 cubic inches and a diameter of 24 inches?
S_A_V [24]
H= .6.5

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3 0
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To rent a uhal truck for the day, it cost 20 dollars per day plus 1 dollar for every mile driven Write an algebraic expression t
valkas [14]

Answer:

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Step-by-step explanation:

To properly represent the algebraic expression, we need to assign some variables.

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R = $1,060

5 0
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