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gulaghasi [49]
3 years ago
10

Write an equation of the line passing through each of the following pairs of points. a (−10, 4), (2, −5)

Mathematics
1 answer:
rusak2 [61]3 years ago
8 0

Answer:

see explanation

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Calculate m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (- 10, 4) and (x₂, y₂ ) = (2, - 5)

m = \frac{-5-4}{2+10} = \frac{-9}{12} = - \frac{3}{4}, thus

y = - \frac{3}{4} x + c ← is the partial equation of the line

To find c substitute either of the 2 given points into the partial equation

Using (2, - 5), then

- 5 = - \frac{3}{2} + c ⇒ c = - 5 + \frac{3}{2} = - \frac{7}{2}

y = - \frac{3}{4} x - \frac{7}{2} ← equation of line

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Levart [38]

A factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

<h3>What are the properties of roots of a polynomial?</h3>
  • The maximum number of roots of a polynomial of degree n is n.
  • For a polynomial with real coefficients, the roots can be real or complex.
  • The complex roots of a polynomial with real coefficients always exist in a pair of conjugate numbers i.e., if a+ib is a root, then a-ib is also a root.

If the roots of the polynomial p(x)=ax^4+bx^3+cx^2+dx+e are r_1,r_2,r_3,r_4, then it can be factorized as p(x)=(x-r_1)(x-r_2)(x-r_3)(x-r_4).

Here, we are to find a factorization of p(x)=x^4+2x^3+7x^2-6x+44. Also, given that -2+i\sqrt{7} and 1-i\sqrt{3} are roots of the polynomial.

Since p(x)=x^4+2x^3+7x^2-6x+44 is a polynomial with real coefficients, so each complex root exists in a pair of conjugates.

Hence, -2-i\sqrt{7} and 1+i\sqrt{3} are also roots of the given polynomial.

Thus, all the four roots of the polynomial p(x)=x^4+2x^3+7x^2-6x+44, are: r_1=-2+i\sqrt{7}, r_2=-2-i\sqrt{7}, r_3=1-i\sqrt{3}, r_4=1+i\sqrt{3}.

So, the polynomial p(x)=x^4+2x^3+7x^2-6x+44 can be factorized as follows:

\{x-(-2+i\sqrt{7})\}\{x-(-2-i\sqrt{7})\}\{x-(1-i\sqrt{3})\}\{x-(1+i\sqrt{3})\}\\=(x+2-i\sqrt{7})(x+2+i\sqrt{7})(x-1+i\sqrt{3})(x-1-i\sqrt{3})\\=\{(x+2)^2+7\}\{(x-1)^2+3\}\hspace{1cm} [\because (a+b)(a-b)=a^2-b^2]\\=(x^2+4x+4+7)(x^2-2x+1+3)\\=(x^2+4x+11)(x^2-2x+4)

Therefore, a factorization of x^4+2x^3+7x^2-6x+44 is (x^2+4x+11)(x^2-2x+4).

To know more about factorization, refer: brainly.com/question/25829061

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Answer:

26x³ - 12x² + 5x + 7

Step-by-step explanation:

  1. (2x³ - 3x + 11) - (3x² + 1) x (4 - 8x)
  2. 2x³ - 3x+11 - (12x² - 24x³ + 4 - 8x)
  3. 2x³ - 3x + 11 - 12x² + 24x³ - 4 + 8x
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= 26x³ - 12x² + 5x + 7

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Identify the range of the function shown in the graph.<br><br> help
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Answer:

-5 ≤ y ≤ 5

Step-by-step explanation:

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