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3 years ago

What is sin θ given (–3, 4) is a point on the terminal side of θ?

Mathematics

1 answer:

Ber [7]3 years ago

5 0

Answer:

QUESTION 1- $sin\theta =-\frac{4}{5}$

QUESTION 2- $cos\theta =-\frac{5}{\sqrt{14} }$

QUESTION3- $sinx =-\frac{1}{\sqrt{17} }$

Step-by-step explanation:

For Question 1

From the given point(-3,4), we can deduce the following.

The point is in the 2nd quadrant.

The right-angled triangle which is formed has a hypotenuse of length 5 by Pythagoras theorem whrere B=3 and P=4.

The third quadrant are from 90° to 180° in which the angle lie.

so $sin\theta =-\frac{P}{H}=-\frac{4}{5}$

For Question 2

From the given point(-4,-5), we can deduce the following.

The point is in the 3rd quadrant.

The right-angled triangle which is formed has a hypotenuse of length $\sqrt{14}$ by Pythagoras theorem whrere B=-4 and P=-5.

The third quadrant are from 180° to 270° in which the angle lie.

so $cos\theta =-\frac{P}{H} =-\frac{5}{\sqrt{14} }$

For Question 3

The line with equation 4b + a = 0 , we can deduce that the value of $b=-\frac{a}{4}$

slope of line $tan\theta =-\frac{P}{B} =-\frac{1}{4}$

Hypotenuse of length $\sqrt{17}$ by Pythagoras theorem whrere B=4and P=1.

so $sinx =-\frac{P}{H} =-\frac{1}{\sqrt{17} }$ .

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miss Akunina [59]

Answer:

Options (1), (2), (3) and (7)

Step-by-step explanation:

Given expression is $\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$ .

Now we will solve this expression with the help of law of exponents.

$\frac{\sqrt[3]{8^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}=\frac{\sqrt[3]{(2^3)^{\frac{1}{3}}\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{\sqrt[3]{2\times 3} }{3\times2^{\frac{1}{9}}}$

$=\frac{2^{\frac{1}{3}}\times 3^{\frac{1}{3}}}{3\times 2^{\frac{1}{9}}}$

$=2^{\frac{1}{3}}\times 3^{\frac{1}{3}}\times 2^{-\frac{1}{9}}\times 3^{-1}$

$=2^{\frac{1}{3}-\frac{1}{9}}\times 3^{\frac{1}{3}-1}$

$=2^{\frac{3-1}{9}}\times 3^{\frac{1-3}{3}}$

$=2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }$ [Option 2]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$ [Option 1]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(\sqrt[9]{2})^2\times (\sqrt[3]{\frac{1}{3} } )^2$

$=(2^2)^{\frac{1}{9}}\times (3^2)^{-\frac{1}{3} }$

$=\sqrt[9]{4}\times \sqrt[3]{\frac{1}{9} }$ [Option 3]

$2^{\frac{2}{9}}\times 3^{-\frac{2}{3} }=(2^2)^{\frac{1}{9}}\times (3^{-2})^{\frac{1}{3} }$

$=\sqrt[9]{2^2}\times \sqrt[3]{3^{-2}}$ [Option 7]

Therefore, Options (1), (2), (3) and (7) are the correct options.

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3 years ago

The midpoint of AB is (-4,-7) if the coordinates of A are (-7,-6) what are the coordinates of B

slava [35]

Answer:

(-1,-8)

Step-by-step explanation:

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3 years ago

Is y=3x+4 parallel or perpendicular

Sauron [17]

Answer: $y=3x+4$ is parallel to $y=3x+7$

Step-by-step explanation:

<h3> The complete exercise is: "Is $y=3x+4$ parallel, perpendicular or neither to $y=3x+7$ ?"</h3><h3 />

The equation of the line in Slope-Intercept form is:

$y=mx+b$

Where "m" is the slope of the line and "b" is the y-intercept.

First, in order to solve this exercise it is important to remember that, by definition:

1. The slopes of parallel lines are equal.

2. The slopes of perpendicular lines are negative reciprocal.

In this case, you have the following line given in the exercise:

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You can identify that "m" and "b" are:

$m=3\\b=4$

And the other line provided in the exercise is this one:

$y=3x+7$

So, you can identify that:

$m=3\\b=7$

As you can notice, the slopes of both lines are equal; therefore, you can conclude that those lines are parallel.

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3 years ago

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iris [78.8K]

Answer:

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