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2 years ago

Use the discriminant to determine the number of real solutions to the quadratic equation.

Mathematics

1 answer:

MAVERICK [17]2 years ago

3 0

Answer:

12x2+12x+1=0

Step-by-step explanation:

well first you need to add all the x's and add your additional numbers, keep all +'s signals and then minuse the addition numbers which makes your anserw negitive and then divide your ansew by your x.

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Jason can travel 24 3/4 miles in 1/2 hour. What is his average speed in miles per hour?

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4) Dan Ariely and colleagues have conducted some extremely interesting studies on cheating (see, for example Mazar, Amir, &

lianna [129]

I've attached the complete question.

Answer:

Only participant 1 is not cheating while the rest are cheating.

Because only participant 1 has a z-score that falls within the 95% confidence interval.

Step-by-step explanation:

We are given;

Mean; μ = 3.3

Standard deviation; s = 1

Participant 1: X = 4

Participant 2: X = 6

Participant 3: X = 7

Participant 4: X = 0

Z - score for participant 1:

z = (x - μ)/s

z = (4 - 3.3)/1

z = 0.7

Z-score for participant 2;

z = (6 - 3.3)/1

z = 2.7

Z-score for participant 3;

z = (7 - 3.3)/1

z = 3.7

Z-score for participant 4;

z = (0 - 3.3)/1

z = -3.3

Now from tables, the z-score value for confidence interval of 95% is between -1.96 and 1.96

Now, from all the participants z-score only participant 1 has a z-score that falls within the 95% confidence interval.

Thus, only participant 1 is not cheating while the rest are cheating.

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3 years ago

Find sinϴ and cosϴ if tanϴ=1/4 and sinϴ>0

eduard

If you're using the app, try seeing this answer through your browser: brainly.com/question/2787701

_______________

$\mathsf{tan\,\theta=\dfrac{1}{4}\qquad\qquad(sin\,\theta\ \textgreater \ 0)}\\\\\\
\mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{1}{4}}\\\\\\
\mathsf{4\,sin\,\theta=cos\,\theta\qquad\quad(i)}$

Square both sides:

$\mathsf{(4\,sin\,\theta)^2=(cos\,\theta)^2}\\\\
\mathsf{4^2\,sin^2\,\theta=cos^2\,\theta}\\\\
\mathsf{16\,sin^2\,\theta=cos^2\,\theta\qquad\qquad(but,~cos^2\,\theta=1-sin^2\,\theta)}\\\\
\mathsf{16\,sin^2\,\theta=1-sin^2\,\theta}$

$\mathsf{16\,sin^2\,\theta+sin^2\,\theta=1}\\\\
\mathsf{17\,sin^2\,\theta=1}\\\\
\mathsf{sin^2\,\theta=\dfrac{1}{17}}\\\\\\
\mathsf{sin\,\theta=\pm\,\sqrt{\dfrac{1}{17}}}\\\\\\
\mathsf{sin\,\theta=\pm\,\dfrac{1}{\sqrt{17}}}$

Since $\mathsf{sin\,\theta}$ is positive, you can discard the negative sign. So,

$\mathsf{sin\,\theta=\dfrac{1}{\sqrt{17}}\qquad\quad\checkmark}$

Substitute this value back into $\mathsf{(i)}$ to find $\mathsf{cos\,\theta:}$

$\mathsf{4\cdot \dfrac{1}{\sqrt{17}}=cos\,\theta}\\\\\\
\mathsf{cos\,\theta=\dfrac{4}{\sqrt{17}}\qquad\quad\checkmark}$

I hope this helps. =)

Tags: <em>trigonometric identity relation trig sine cosine tangent sin cos tan trigonometry precalculus</em>

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3 years ago

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