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3 years ago

If sin theta= 2/3 and tan theta<0, what is the value of cos theta

2 answers:

8 0

Cos
θ
=
√
5
3

or it could be cos
θ
=
√
5
−
3

Explanation:
Since sin
θ
is negative, it can be in the third or fourth quadrant
Drawing your right-angled triangle, place your
θ
in one of three corners. Your longest side will be 3 and the side opposite the
θ
will be -2. Finally, using Pythagoras theorem, your last side should be
√
5

Now, if your triangle was in the third quadrant, you would have
cos
θ
=
√
5
−
3
since cosine is negative in the third quadrant
But if your triangle was in the fourth quadrant, you would have
cos
θ
=
√
5
3
since cosine is positive in the fourth quadrant

PilotLPTM [1.2K]3 years ago

6 0

Answer: - sqrt 5 / 3

Step-by-step explanation:

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-3(t+5)+(4t+2)=8

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First use distributive property on -3(t+5), which would be -3t minus -15.

Then, add all like terms: -3t-15+4t+2 ---> -3t+4t+2-15 --> t-13.

t-13=8, add 13 to both sides. This cancels the -13 and makes 8 to 21.

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3 years ago

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How many subsets of at least one element does a set of seven elements have?

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3 years ago

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3 years ago

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Find the equation of the sphere if one of its diameters has endpoints (4, 2, -9) and (6, 6, -3) which has been normalized so tha

Pavel [41]

Answer:

$(x - 5)^2 + (y - 4)^2 + (z - 6)^2 = 14$ .

(Expand to obtain an equivalent expression for the sphere: $x^2 - 10\,x + y^2 - 8\, y + z^2 - 12\, z + 63 = 0$ )

Step-by-step explanation:

Apply the Pythagorean Theorem to find the distance between these two endpoints:

$\begin{aligned}&\text{Distance}\cr &= \sqrt{\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2 + \left(z_2 - z_1\right)^2} \cr &= \sqrt{(6 - 4)^2 + (6 - 2)^2 + ((-3) - (-9))^2 \cr &= \sqrt{56}}\end{aligned}$ .

Since the two endpoints form a diameter of the sphere, the distance between them would be equal to the diameter of the sphere. The radius of a sphere is one-half of its diameter. In this case, that would be equal to:

$\begin{aligned} r &= \frac{1}{2} \, \sqrt{56} \cr &= \sqrt{\left(\frac{1}{2}\right)^2 \times 56} \cr &= \sqrt{\frac{1}{4} \times 56} \cr &= \sqrt{14} \end{aligned}$ .

In a sphere, the midpoint of every diameter would be the center of the sphere. Each component of the midpoint of a segment (such as the diameter in this question) is equal to the arithmetic mean of that component of the two endpoints. In other words, the midpoint of a segment between $\left(x_1, \, y_1, \, z_1\right)$ and $\left(x_2, \, y_2, \, z_2\right)$ would be:

$\displaystyle \left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right)$ .

In this case, the midpoint of the diameter, which is the same as the center of the sphere, would be at:

$\begin{aligned}&\left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right) \cr &= \left(\frac{4 + 6}{2},\, \frac{2 + 6}{2}, \, \frac{(-9) + (-3)}{2}\right) \cr &= (5,\, 4\, -6)\end{aligned}$ .

The equation for a sphere of radius $r$ and center $\left(x_0,\, y_0,\, z_0\right)$ would be:

$\left(x - x_0\right)^2 + \left(y - y_0\right)^2 + \left(z - z_0\right)^2 = r^2$ .

In this case, the equation would be:

$\left(x - 5\right)^2 + \left(y - 4\right)^2 + \left(z - (-6)\right)^2 = \left(\sqrt{56}\right)^2$ .

Simplify to obtain:

$\left(x - 5\right)^2 + \left(y - 4\right)^2 + \left(z + 6\right)^2 = 56$ .

Expand the squares and simplify to obtain:

$x^2 - 10\,x + y^2 - 8\, y + z^2 - 12\, z + 63 = 0$ .

8 0

3 years ago

Points F,G, and H are collinear such that FG:FH=2/3.If a directed line segment begins at F(-3,2) and ends at H (-3,7), find the

Nana76 [90]

We consider the x- and y-coordinates separately. Let the coordinates of G be (x, y). Now considering the x-coordinates:
FG/FH = (x - (-3)) / (-3 - (-3)) = 2/3
x + 3 = (2/3)(6)
x = 1
For the y-coordinates:
FG/FH = (y - 2) / (7 - 2) = 2/3
y - 2 = (2/3)(5)
y = 16/3
Therefore the coordinates of G are (1, 16/3).

8 0

3 years ago