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vfiekz [6]
3 years ago
11

Which equation best represents the line? (10 Points)

Mathematics
1 answer:
son4ous [18]3 years ago
7 0

Answer: D. y = 3x − 1

Step-by-step explanation:

The equation of a straight line can be represented in the slope-intercept form, y = mx + c

Where c = intercept

Slope, m =change in value of y on the vertical axis / change in value of x on the horizontal axis

change in the value of y = y2 - y1

Change in value of x = x2 -x1

y2 = final value of y

y 1 = initial value of y

x2 = final value of x

x1 = initial value of x

From the graph,

y2 = 2

y1 = - 1

x2 = 1

x1 = 0

Slope,m = (2 - - 1)/(1 - 0) = 3/1 = 3

To determine the intercept, we would substitute x = 1, y = 2 and m= 3 into y = mx + c

y = mx + c. It becomes

2 = 3 × 1 + c = 3 + c

c = 2 - 3 = - 1

The equation becomes

y = 3x - 1

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In a stack of boards at a lumber yard, the 20th board counting from the top of the stack is immediately below the 16th board cou
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Answer:

C

Step-by-step explanation:

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The value of -4/5 + 9/20 is
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-4/5+9/20=-16/20+9/20=-7/20
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2 years ago
_ 5x^20-7x^10+15 in quadratic form
Tpy6a [65]

Answer:

x = -(sqrt(349) - 7)^(1/10)/10^(1/10) or x = (sqrt(349) - 7)^(1/10)/10^(1/10) or x = -((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -(1/10 (-7 - sqrt(349)))^(1/10) or x = (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(1/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(1/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(2/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(2/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(3/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(3/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(4/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(4/5) (1/10 (-7 - sqrt(349)))^(1/10)

Step-by-step explanation:

Solve for x:

-5 x^20 - 7 x^10 + 15 = 0

Substitute y = x^10:

-5 y^2 - 7 y + 15 = 0

Divide both sides by -5:

y^2 + (7 y)/5 - 3 = 0

Add 3 to both sides:

y^2 + (7 y)/5 = 3

Add 49/100 to both sides:

y^2 + (7 y)/5 + 49/100 = 349/100

Write the left hand side as a square:

(y + 7/10)^2 = 349/100

Take the square root of both sides:

y + 7/10 = sqrt(349)/10 or y + 7/10 = -sqrt(349)/10

Subtract 7/10 from both sides:

y = sqrt(349)/10 - 7/10 or y + 7/10 = -sqrt(349)/10

Substitute back for y = x^10:

x^10 = sqrt(349)/10 - 7/10 or y + 7/10 = -sqrt(349)/10

Taking 10^th roots gives (sqrt(349)/10 - 7/10)^(1/10) times the 10^th roots of unity:

x = -(1/10 (sqrt(349) - 7))^(1/10) or x = (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(1/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(1/5) (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(2/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(2/5) (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(3/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(3/5) (1/10 (sqrt(349) - 7))^(1/10) or x = -(-1)^(4/5) (1/10 (sqrt(349) - 7))^(1/10) or x = (-1)^(4/5) (1/10 (sqrt(349) - 7))^(1/10) or y + 7/10 = -sqrt(349)/10

Subtract 7/10 from both sides:

x = -(sqrt(349) - 7)^(1/10)/10^(1/10) or x = (sqrt(349) - 7)^(1/10)/10^(1/10) or x = -((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or y = -7/10 - sqrt(349)/10

Substitute back for y = x^10:

x = -(sqrt(349) - 7)^(1/10)/10^(1/10) or x = (sqrt(349) - 7)^(1/10)/10^(1/10) or x = -((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x^10 = -7/10 - sqrt(349)/10

Taking 10^th roots gives (-7/10 - sqrt(349)/10)^(1/10) times the 10^th roots of unity:

Answer:  x = -(sqrt(349) - 7)^(1/10)/10^(1/10) or x = (sqrt(349) - 7)^(1/10)/10^(1/10) or x = -((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(1/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(2/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(3/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = ((-1)^(4/5) (sqrt(349) - 7)^(1/10))/10^(1/10) or x = -(1/10 (-7 - sqrt(349)))^(1/10) or x = (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(1/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(1/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(2/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(2/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(3/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(3/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = -(-1)^(4/5) (1/10 (-7 - sqrt(349)))^(1/10) or x = (-1)^(4/5) (1/10 (-7 - sqrt(349)))^(1/10)

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(6+√-64)(3-√-16) Multiply. This is urgent, please explain this to me!
romanna [79]

Answer:

50

Step-by-step explanation:

\sqrt{-64} and \sqrt{-16} can be expressed in complex form, with \sqrt{-1} = i

\sqrt{-64} = \sqrt{64(-1)} = \sqrt{64} × \sqrt{-1} = 8i

\sqrt{-16} = \sqrt{16(-1)} = \sqrt{16} × \sqrt{-1} = 4i

the factors can then be expressed as

(6 + 8i)(3 - 4i) ← expand using FOIL

= 18 - 24i + 24i - 32i² [ i² = (\sqrt{-1} )² = - 1 ]

= 18 - 24i + 24i + 32 ← collect like terms

= 18 + 32 + 0

= 50

6 0
1 year ago
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