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Temka [501]
3 years ago
12

Graph the linear equation. Find three points that solve the equation, then plot on the graph. -x - 2y = -10

Mathematics
1 answer:
jok3333 [9.3K]3 years ago
3 0
-x - 2y = -10  |add x to both sides

-2y = x - 10    |divide both sides by (-2)

y = -0.5x + 5

for x = 0, y = -0.5(0) + 5 = 0 + 5 = 5 ⇒ (0; 5)
for x = -2, y = -0.5(-2) + 5 = 1 + 5 = 6 ⇒ (-2; 6)
for x = 2, y= -0.5(2) + 5 = -1 + 5 = 4 ⇒ (2; 4)

The graph in the picture

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PLEASE HELP!!!! Kelly earns 20% more than Scott, and together they earn $1210.00 per week. How much does Kelly earn per week?
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Answer:

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=242

Kelly earns

242+1210

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3 0
3 years ago
Divide 120 in the ratio of 1:2
antiseptic1488 [7]

Ratio given 1:2

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40 and 80 are in the ratio 1:2 of 120.

Hope it helps!!!

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Explain how you would use properties do you write an expression equivalent to 7y+4b-3y.
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5 0
3 years ago
Match the hyperbolas represented by the equations to their foci.
Arte-miy333 [17]

Answer:

1) (1 , -22) and (1 , 12) ⇔ (y + 5)²/15² - (x - 1)²/8² = 1

2) (-7 , 5) and (3 , 5) ⇔ (x + 2)²/3² - (y - 5)²/4² = 1

3) (-6 , -2) and (14 , -2) ⇔ (x - 4)²/8² - (y + 2)²/6² = 1

4) (-7 , -10) and (-7 , 16) ⇔ (y - 3)²/5² - (x + 7)²/12² = 1

Step-by-step explanation:

* Lets study the equation of the hyperbola

- The standard form of the equation of a hyperbola with

  center (h , k) and transverse axis parallel to the x-axis is

  (x - h)²/a² - (y - k)²/b² = 1

- the coordinates of the foci are (h ± c , k), where c² = a² + b²

- The standard form of the equation of a hyperbola with

  center (h , k) and transverse axis parallel to the y-axis is

  (y - k)²/a² - (x - h)²/b² = 1

- the coordinates of the foci are (h , k ± c), where c² = a² + b²

* Lets look to the problem

1) The foci are (1 , -22) and (1 , 12)

- Compare the point with the previous rules

∵ h = 1 and k ± c = -22 ,12

∴ The form of the equation will be (y - k)²/a² - (x - h)²/b² = 1

∵ k + c = -22 ⇒ (1)

∵ k - c = 12 ⇒ (2)

* Add (1) and(2)

∴ 2k = -10 ⇒ ÷2

∴ k = -5

* substitute the value of k in (1)

∴ -5 + c = -22 ⇒ add 5 to both sides

∴ c = -17

∴ c² = (-17)² = 289

∵ c² = a² + b²

∴ a² + b² = 289

* Now lets check which answer has (h , k) = (1 , -5)

  and a² + b² = 289 in the form (y - k)²/a² - (x - h)²/b² = 1

∵ 15² + 8² = 289

∵ (h , k) = (1 , -5)

∴ The answer is (y + 5)²/15² - (x - 1)²/8² = 1

* (1 , -22) and (1 , 12) ⇔ (y + 5)²/15² - (x - 1)²/8² = 1

2) The foci are (-7 , 5) and (3 , 5)

- Compare the point with the previous rules

∵ k = 5 and h ± c = -7 ,3

∴ The form of the equation will be (x - h)²/a² - (y - k)²/b² = 1

∵ h + c = -7 ⇒ (1)

∵ h - c = 3 ⇒ (2)

* Add (1) and(2)

∴ 2h = -4 ⇒ ÷2

∴ h = -2

* substitute the value of h in (1)

∴ -2 + c = -7 ⇒ add 2 to both sides

∴ c = -5

∴ c² = (-5)² = 25

∵ c² = a² + b²

∴ a² + b² = 25

* Now lets check which answer has (h , k) = (-2 , 5)

  and a² + b² = 25 in the form (x - h)²/a² - (y - k)²/b² = 1

∵ 3² + 4² = 25

∵ (h , k) = (-2 , 5)

∴ The answer is (x + 2)²/3² - (y - 5)²/4² = 1

* (-7 , 5) and (3 , 5) ⇔ (x + 2)²/3² - (y - 5)²/4² = 1

3) The foci are (-6 , -2) and (14 , -2)

- Compare the point with the previous rules

∵ k = -2 and h ± c = -6 ,14

∴ The form of the equation will be (x - h)²/a² - (y - k)²/b² = 1

∵ h + c = -6 ⇒ (1)

∵ h - c = 14 ⇒ (2)

* Add (1) and(2)

∴ 2h = 8 ⇒ ÷2

∴ h = 4

* substitute the value of h in (1)

∴ 4 + c = -6 ⇒ subtract 4 from both sides

∴ c = -10

∴ c² = (-10)² = 100

∵ c² = a² + b²

∴ a² + b² = 100

* Now lets check which answer has (h , k) = (4 , -2)

  and a² + b² = 100 in the form (x - h)²/a² - (y - k)²/b² = 1

∵ 8² + 6² = 100

∵ (h , k) = (4 , -2)

∴ The answer is (x - 4)²/8² - (y + 2)²/6² = 1

* (-6 , -2) and (14 , -2) ⇔ (x - 4)²/8² - (y + 2)²/6² = 1

4) The foci are (-7 , -10) and (-7 , 16)

- Compare the point with the previous rules

∵ h = -7 and k ± c = -10 , 16

∴ The form of the equation will be (y - k)²/a² - (x - h)²/b² = 1

∵ k + c = -10 ⇒ (1)

∵ k - c = 16 ⇒ (2)

* Add (1) and(2)

∴ 2k = 6 ⇒ ÷2

∴ k = 3

* substitute the value of k in (1)

∴ 3 + c = -10 ⇒ subtract 3 from both sides

∴ c = -13

∴ c² = (-13)² = 169

∵ c² = a² + b²

∴ a² + b² = 169

* Now lets check which answer has (h , k) = (-7 , 3)

  and a² + b² = 169 in the form (y - k)²/a² - (x - h)²/b² = 1

∵ 5² + 12² = 169

∵ (h , k) = (-7 , 3)

∴ The answer is (y - 3)²/5² - (x + 7)²/12² = 1

* (-7 , -10) and (-7 , 16) ⇔ (y - 3)²/5² - (x + 7)²/12² = 1

7 0
3 years ago
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