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3 years ago

Solve by completing the square.

Mathematics

2 answers:

IRINA_888 [86]3 years ago

7 0

Answer:

-23.18, -0.82

Step-by-step explanation:

h² + 24h + 19 = 0
h²+2h*12+12²-125=0
(h+12)²-√125²=0
(h+12+11.18)(h+12-11.18)=0
(h+23.18)(h+0.82)=0
h+23.18=0 ⇒ h= -23.18
h+0.82=0 ⇒ h= -0.82

Colt1911 [192]3 years ago

7 0

Answer:

-23.18, -.82

Step-by-step explanation:

h^2 + 24h + 19 = 0

Subtract 19 from each side

h^2 +24h = -19

Take the coefficient of h

Divide by 2

24/2 =12

Square it

12^2 =144

Add this to both sides

h^2 +24h+144 = -19+144

(h+12)^2 =125

Take the square root of each side

sqrt((h+12)^2) =±sqrt(125)

h+12 = ±sqrt(125)

Subtract 12 from each side

h+12-12 = -12 ±sqrt(125)

h =-12 ±sqrt(125)

h = -12 - sqrt(125) = -23.18033989

h = -12 + sqrt(125) = -.819660113

To the nearest hundredth

-23.18, -.82

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Answer:

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-7=n-6

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3 years ago

How do you solve this problem?

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3 years ago

Which table and graph represent the equation y = 5x? A. Table B and graph B B. Table A and graph A C. Table A and graph B D. Tab

kaheart [24]

Answer:

Summary:

We conclude that:

Table B and Graph A represents the equation.

Hence, option D represents the equation.

Step-by-step explanation:

Given the equation

y = 5x

CHECKING TABLE A

x y

5 1

10 2

15 3

substituting x = 5 in the equation

y = 5x

y = 5(5)

y = 25

Thus,

at x = 5, y = 25

Hence,

(5, 25) does not satisfy the table A.

substituting x = 10 in the equation

y = 5x

y = 5(10)

y = 50

Thus,

at x = 10, y = 50

Hence,

(10, 50) does not satisfy the table A.

substituting x = 15 in the equation

y = 5x

y = 5(15)

y = 75

Thus,

at x = 15, y = 75

Hence,

(15, 75) does not satisfy the table A.

Conclusion:

Table A does not represent the equation y = 5x

CHECKING TABLE B

x y

1 5

2 10

3 15

substituting x = 1 in the equation

y = 5x

y = 5(1)

y = 5

Thus,

at x = 1, y = 5

Hence,

(1, 5) satisfies the table B.

substituting x = 2 in the equation

y = 5x

y = 5(2)

y = 10

Thus,

at x = 2, y = 10

Hence,

(2, 10) satisfies the table B.

substituting x = 3 in the equation

y = 5x

y = 5(3)

y = 15

Thus,

at x = 3, y = 15

Hence,

(3, 15) satisfies the table B.

Conclusion:

Table A represents the equation y = 5x

CHECKING GRAPH A

Given the equation

y = 5x

It is clear from graph A that

at x = 1, y = 5

at x = 2, y = 10

at x = 3, y = 15

Also

Putting x = 0,

y = 5(0) = 0

Thus, at x = 0, y = 0

Aso the table indicates that at x = 0, y = 0

Thus, the graph represents the equation y = 5x.

CHECKING GRAPH B

y = 5x

Putting x = 0,

y = 5(0) = 0

Thus, at x = 0, y = 0

But, the table B indicates that at x = 0, y = 5.

Thus, graph B does not represent the equation

Summary:

We conclude that:

Table B and Graph A represents the equation.

Hence, option D represents the equation.

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3 years ago

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On highway 77, the Dobson exit is halfway between the Mt. Airy exit and the Elkin exit. If the Elkin exit is located at (-5,-2)

Leya [2.2K]

Given :

The Dobson exit is halfway between the Mt. Airy exit and the Elkin exit. If the Elkin exit is located at (-5,-2) and the Dobson exit is located at (-1,4).

To Find :

How far is it from the Elkin exit to the Mt. Airy exit.

Solution :

E(-5,-2) , D(-1 ,4 ) .

It is also given that Dobson exit is halfway between the Mt. Airy exit and the Elkin exit.

Let , location of Elkin exit is A(h,k) .

So , point D in terms of A and E is given by :

$(\dfrac{h-5}{2},\dfrac{k-2}{2})$ .

Comparing it with given D :

$\dfrac{h-5}{2}=-1\\\\h=3$ $\dfrac{k-2}{2}=4\\\\k=10$

Therefore , the location of Mt. Airy exit is ( 3, 10 ) .

Distance between Elkin exit to the Mt. Airy exit.

$D=\sqrt{(5-3)^2+(-2-10)^2}\\\\D=12.17\ units$

Therefore , distance between Elkin exit to the Mt. Airy exit is 12.17 units .

Hence ,this is the required solution .

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3 years ago