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3 years ago

Select the polynomial that is a perfect square trinomial.

Mathematics

2 answers:

Sveta_85 [38]3 years ago

7 0

Answer:

The answer is 16x2 − 8x + 1

To be a perfect square trinomial, the rule must be used:

(a - b)² = a² - 2ab + b²

Let's check all choices

A. a = 7x, b = 4

(7x - 4)² = 49x² - 56x + 16

B. a = 2a, b = 5

(2a - 5)² = 4a² - 20a + 25

C. a = 5b, b = √10

(5b - √10)² = 25b² - 10√10 b + 10

D. a = 4x, b = 1

(4x - 1)² = 16x² - 8x + 1

CORRECT

Step-by-step explanation:

cestrela7 [59]3 years ago

3 0

The answer is <span>16x2 − 8x + 1
</span>
To be a perfect square trinomial, the rule must be used:
(a - b)² = a² - 2ab + b²

Let's check all choices
A. a = 7x, b = 4
(7x - 4)² = 49x² - 56x + 16

B. a = 2a, b = 5
(2a - 5)² = 4a² - 20a + 25

C. a = 5b, b = √10
(5b - √10)² = 25b² - 10√10 b + 10

D. a = 4x, b = 1
(4x - 1)² = 16x² - 8x + 1
CORRECT

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Answer:

The triangle is both an Isosceles triangle and a right triangle.

Step-by-step explanation:

Given the vertices of a triangle.

$P_{1} = (- 1, 4)$

$P_{2} = (6, 2)$ and

$P_{3} = (4, - 5)$

We find the distance between all the points to determine the length of each side of the triangle.

Distance between any two points, say, $(x_1, y_1)$ and $(x_2, y_2)$ is:

$\sqrt{\bigg ( \textbf{x}_{\textbf{2}} \hspace{1mm} \textbf{- x}_{\textbf{1}} \bigg )^{\textbf{2}} \textbf{+} \bigg( \textbf{y}_{\textbf{2}} \hspace{1mm} \textbf{- y}_{\textbf{1}} \bigg)^ {\textbf{2}}$

Length between $P_1$ and $P_{2}$ , (Side 1) :

$(x_1, y_1) = (- 1, 4)$ and

$(x_2, y_2) = (6, 2)$

Distance = $\sqrt{\bigg(6 - (-1) \bigg)^{2} \hspace{1mm} + \hspace{1mm} \bigg( 2 - 4 \bigg )^2$

$= \sqrt{7^2 + 2^2}$

$= \sqrt{49 + 4}$

$= \sqrt{\textbf{53}}$

Length of Side 1 = $\sqrt{\textbf{53}}$ units.

Distance between $P_1$ and $P_2$ , (Side 2):

$(x_1, y_1) = (-1, 4)$

$(x_2, y_2) = (4, - 5)$

Distance = $\sqrt{ \bigg( 4 + 1 \bigg)^2 \hspace{1mm} + \bigg( - 5 - 4 \bigg ) ^2$

$= \sqrt{ 25 + 81 }$

$= \sqrt{\textbf{106}}$

Length of Side 2 = $\sqrt{\textbf{106}}$ units.

Distance between $P_2$ and $P_3$ , Side 3 :

$(x_1, y_1) = (4, 5)$

$(x_2, y_2) = (6, 2)$

Distance = $\sqrt{ 2^2 \hspace{1mm} + \hspace{1mm} 7^2}$

$= \sqrt{49 + 4}$

$= \sqrt{\textbf{53}$

Length of Side 3 = $\sqrt{\textbf{53}}$ units.

Note that the length of Side 1 = Length of Side 3.

That means the triangle is Isosceles.

Also, For a triangle to be right angle triangle, using Pythagoras theorem we have:

(Side 1)² + (Side 3)² = (Side 2)²

$\bigg( \sqrt{53} \bigg )^2 \hspace{1mm} + \hspace{1mm} \bigg( \sqrt{53} \bigg)^2 \hspace{1mm} = \hspace{1mm} \bigg ( \sqrt{106} \bigg ) ^2$

i.e, 53 + 53 = 106

Hence, the triangle is a right - angled triangle as well.

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