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olya-2409 [2.1K]
3 years ago
8

Theta is an acute angle and sin theta and cos theta are given. Use identities to find tan, csc, sec, and cot. Where necessary, r

ationalize denominators. Sin theta = 8/17, cos theta = 15/17.
a. tan theta = ?
b. csc theta = ?
c. sec theta = ?
d. cot theta = ?
Mathematics
1 answer:
dimulka [17.4K]3 years ago
5 0

Answer: C. sec theta

You might be interested in
The length of a rectangle is three times it's width. If the width is diminished by 1 m and length is increased by 3 m, the area
Stella [2.4K]

Answer: the length is 15 m and the width is 5 m

Step-by-step explanation:

Let L represent the original length of the rectangle.

Let W represent the original width of the rectangle.

The length of a rectangle is three times it's width. This means that

L = 3W

If the width is diminished by 1 m and length is increased by 3 m, the area of the rectangle that is formed is 72 m². This means that

(L + 3)(W - 1) = 72

LW - L + 3W - 3 - 72 = 72 + 3

LW - L + 3W = 75 - - - - -- - - - - - 1

Substituting L = 3W into equation 1, it becomes

3W × W - 3W + 3W = 75

3W² = 75

W² = 75/3 = 25

W = √25 = ±5

The width cannot be negative. Therefore, the width is 5 m

L = 3W = 3 × 5 = 15 m

3 0
3 years ago
Write the equation of the line that is perpendicular to the line y = 2x + 2 and passes through the point (6, 3).
sdas [7]

There asking for the equation, and the possible answers are 

y = 2x + 6 y = −one halfx + 3 y = −one halfx + 6 y = 2x + 3


4 0
2 years ago
What is d divided by 2
beks73 [17]

Answer:

d / 2

Step-by-step explanation:

6 0
2 years ago
Daisy taped two pieces of cardboard together to make a longer piece. The length of the first piece of cardboard was 1 2/5 meters
N76 [4]
T= total length

T=1 2/5 meters + 4 1/5 meters
add the whole #s and the fractions
T= (1+4) + (2/5 + 1/5)
T= 5 3/5 meters total

Hope this helps! :)


7 0
3 years ago
Find the standard form of the equation for the conic section represented by x^2 + 10x + 6y = 47.
Levart [38]

Answer:

The standard form of the equation for the conic section represented by x^2\:+\:10x\:+\:6y\:=\:47 is:

4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2

Step-by-step explanation:

We know that:

4p\left(y-k\right)=\left(x-h\right)^2 is the standard equation for an up-down facing Parabola with vertex at (h, k), and focal length |p|.

Given the equation

x^2\:+\:10x\:+\:6y\:=\:47

Rewriting the equation in the standard form

4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2

Thus,

The vertex (h, k) = (-5, 12)

Please also check the attached graph.

Therefore, the standard form of the equation for the conic section represented by x^2\:+\:10x\:+\:6y\:=\:47 is:

4\left(-\frac{3}{2}\right)\left(y-12\right)=\left(x-\left(-5\right)\right)^2

where

vertex (h, k) = (-5, 12)

7 0
2 years ago
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