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Anettt [7]
3 years ago
14

Use the zero product property to find the solutions to the equation 2x2 + x - 1 = 2

Mathematics
1 answer:
Firdavs [7]3 years ago
8 0

Answer:

C

Step-by-step explanation:

Given

2x² + x - 1 = 2 ( subtract 2 from both sides )

2x² + x - 3 = 0

Consider the factors of the product of the coefficient of the x² term and the constant term which sum to give the coefficient of the x- term.

product = 2 × - 3 = - 6 and sum = + 1

The factors are - 2 and + 3

Use these factors to split the x- term

2x² - 2x + 3x - 3 = 0 ( factor the first/second and third/fourth terms )

2x(x - 1) + 3(x - 1) = 0 ← factor out (x - 1) from each term

(x - 1)(2x + 3) = 0

Equate each factor to zero and solve for x

x - 1 = 0 ⇒ x = 1

2x + 3 = 0 ⇒ 2x = - 3 ⇒ x = - \frac{3}{2}

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Answer: Options 1, 3, 5

Step-by-step explanation:

Perpendicular lines have slopes that are negative reciprocals, so since the slope of the given line is 1/3, we need to find lines with a slope of -3.

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How do you solve this?
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3 years ago
Read 2 more answers
HELP! Find the value of sin 0 if tan 0 = 4; 180 < 0< 270
BabaBlast [244]

Hi there! Use the following identities below to help with your problem.

\large \boxed{sin \theta = tan \theta cos \theta} \\  \large \boxed{tan^{2}  \theta + 1 =  {sec}^{2} \theta}

What we know is our tangent value. We are going to use the tan²θ+1 = sec²θ to find the value of cosθ. Substitute tanθ = 4 in the second identity.

\large{ {4}^{2}  + 1 =  {sec}^{2} \theta } \\  \large{16 + 1 =  {sec}^{2} \theta } \\  \large{ {sec}^{2}  \theta = 17}

As we know, sec²θ = 1/cos²θ.

\large \boxed{sec \theta =   \frac{1}{cos \theta} } \\  \large \boxed{ {sec}^{2}  \theta =  \frac{1}{ {cos}^{2}  \theta} }

And thus,

\large{  {cos}^{2}  \theta =  \frac{1}{17}}   \\ \large{cos \theta =  \frac{ \sqrt{1} }{ \sqrt{17} } } \\  \large{cos \theta =  \frac{1}{ \sqrt{17} }  \longrightarrow  \frac{ \sqrt{17} }{17} }

Since the given domain is 180° < θ < 360°. Thus, the cosθ < 0.

\large{cos \theta =   \cancel\frac{ \sqrt{17} }{17} \longrightarrow cos \theta =  -  \frac{ \sqrt{17} }{17}}

Then use the Identity of sinθ = tanθcosθ to find the sinθ.

\large{sin \theta = 4 \times ( -  \frac{ \sqrt{17} }{17}) } \\  \large{sin \theta =  -  \frac{4 \sqrt{17} }{17} }

Answer

  • sinθ = -4sqrt(17)/17 or A choice.
4 0
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