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nikdorinn [45]
4 years ago
15

(-9, -10) through the origin

Mathematics
1 answer:
Anit [1.1K]4 years ago
6 0
So (x, y) and origin is is (0, 0) then you place three points on the graph, they should be as follows [(-9, -10), (0, 0), (9, 10)]. Count the spaces you moved from point a to point b, should be right 9 and up 10. meaning...
Slope = 9/10
I'm pretty sure this is correct, I hope this helped.
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Step-by-step explanation:

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Vladimir [108]
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Solve the following system of equations for x to the nearest hundredth : y + 2x + 1 = 0; 4y - 4x ^ 2 - 12x = - 7
Sever21 [200]

Answer:

+3.464; -3.464

Step-by-step explanation:

call A = y + 2x + 1 = 0 => y = (1 - 2x)

call B: 4y - 4(x^2) - 12x = -7

=> replace y from A to B =>

  1. 4(1 - 2x) - 4(x^2) - 12x = -7
  2. 4 - 8x - 4(x ^ 2) - 12x = -7
  3. -8x - 4(x ^ 2) - 12x = -7 - 4 = -11
  4. -4(x^2) - (8x - 12x) = -11
  5. -4(x^2) + 4x = -11
  6. -4(x^2) + 4x + 11 = 0

=> get delta Δ = (-4^2) - 4*(-4 * 11) = 192

=> Δ > 0 => got 2 No

=> x1 = \frac{-4 + \sqrt{192} }{2 * -4} = \frac{1 - 2\sqrt{3} }{2} = -1.232

=> x2 = \frac{-4 - \sqrt{192} }{2 * -4}=\frac{1 + 2\sqrt{3} }{2}= 2.232

=> replace x from B into A

=> y1 = (1 - 2x) = (1 - 2 * -1.232) = 3.464

=> y2 = (1 - 2x) = (1 - 2 * 2.232) = - 3.464

7 0
3 years ago
What is the value of the leading coefficient a, if the polynomial function P(x) = a(x + b)2(x − c) has a multiplicity of 2 at th
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If the polynomial function P(x) = a(x + b)^2(x-c) has a multiplicity of 2 at the point (−1, 0) then the factor (x-(-1))=(x+1) twice enters the polynomial representation. So, you have known one part of the left side of the polynomial function that is P(x) = a(x + 1)^2(x-c).

If polynomial function passes through the point (7,0), then 0 = a(7+ 1)^2(7-c) and since a\neq 0 you have that c=7.

At last, if polynomial function passes through the point (0,-14), then -14 = a(0+ 1)^2(0-7) and -14=-7a, so a=2.
Answer: a=2.



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Solve for x:<br><br> 5/2 = x/(1/7)
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X=5/14 this it in sure
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