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3 years ago

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12.) Given the following information, determine which lines, if any, are parallel. State the converse that justifies your answer

. Pleaseee helppp

1 answer:

Maru [420]3 years ago

6 0

Answer:

Step-by-step explanation:

Given Parallel lines Converse

a. ∠13 ≅ ∠17 a║c Corresponding angles

b. ∠4 ≅ ∠9 d║e Exterior alternate angles

c. m∠20 + m∠21 = 180° a║b Consecutive interior angles

d. ∠8 ≅ ∠19 Vertical angles

e. ∠10 ≅ ∠23 b║c Interior alternate angles

f. m∠14 + m∠17 = 180° a║c Consecutive interior angles

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The graph of a quadratic function has roots at -3 and -7 and a vertex at (-5,4). What is the

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The equation of the function in vertex form is y = -(x + 5)² + 4

<h3>What is quadratic equation?</h3>

A quadratic equation is a second-order polynomial equation in a single variable x , ax2+bx+c=0. with a ≠ 0 .

Given that the quadratic function has roots at -3 and -7 and a vertex at (-5,4).

We need to find the equation of the function in vertex form

equation of the function in vertex form.

As per the information given in the question,

The given roots of the function are -3 and -7,

y = a (x + 3) (x + 7)

y = a (x² + 10x + 21)

y = a (x² + 10x + 25 − 4)

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y = -(x + 5)² + 4

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1 year ago

(a) Determine a cubic polynomial with integer coefficients which has $\sqrt[3]{2} + \sqrt[3]{4}$ as a root.

Keith_Richards [23]

Answer:

(a) $x\³ - 6x - 6$

(b) Proved

Step-by-step explanation:

Given

$r = $\sqrt[3]{2} + \sqrt[3]{4}$$ --- the root

Solving (a): The polynomial

A cubic function is represented as:

$f = (a + b)^3$

Expand

$f = a^3 + 3a^2b + 3ab^2 + b^3$

Rewrite as:

$f = a^3 + 3ab(a + b) + b^3$

The root is represented as:

$r=a+b$

By comparison:

$a = $\sqrt[3]{2}$

$b = \sqrt[3]{4}$$

So, we have:

$f = ($\sqrt[3]{2})^3 + 3*$\sqrt[3]{2}*\sqrt[3]{4}$*($\sqrt[3]{2} + \sqrt[3]{4}$) + (\sqrt[3]{4}$)^3$

Expand

$f = 2 + 3*$\sqrt[3]{2*4}*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

$f = 2 + 3*$\sqrt[3]{8}*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

$f = 2 + 3*2*($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

$f = 2 + 6($\sqrt[3]{2} + \sqrt[3]{4}$) + 4$

Evaluate like terms

$f = 6 + 6($\sqrt[3]{2} + \sqrt[3]{4}$)$

Recall that: $r = $\sqrt[3]{2} + \sqrt[3]{4}$$

So, we have:

$f = 6 + 6r$

Equate to 0

$f - 6 - 6r = 0$

Rewrite as:

$f - 6r - 6 = 0$

Express as a cubic function

$x^3 - 6x - 6 = 0$

Hence, the cubic polynomial is:

$f(x) = x^3 - 6x - 6$

Solving (b): Prove that r is irrational

The constant term of $x^3 - 6x - 6 = 0$ is -6

The divisors of -6 are: -6,-3,-2,-1,1,2,3,6

Calculate f(x) for each of the above values to calculate the remainder when f(x) is divided by any of the above values

$f(-6) = (-6)^3 - 6*-6 - 6 = -186$

$f(-3) = (-3)^3 - 6*-3 - 6 = -15$

$f(-2) = (-2)^3 - 6*-2 - 6 = -2$

$f(-1) = (-1)^3 - 6*-1 - 6 = -1$

$f(1) = (1)^3 - 6*1 - 6 = -11$

$f(2) = (2)^3 - 6*2 - 6 = -10$

$f(3) = (3)^3 - 6*3 - 6 = 3$

$f(6) = (6)^3 - 6*6 - 6 = 174$

For r to be rational;

The divisors of -6 must divide f(x) without remainder

i.e. Any of the above values must equal 0

<em>Since none equals 0, then r is irrational</em>

3 0

2 years ago