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Bumek [7]
2 years ago
15

The statement tan theta -12/5, csc theta -13/5, and the terminal point determined by theta is in quadrant 2."​

Mathematics
1 answer:
Harlamova29_29 [7]2 years ago
6 0

Answer:

Answer C:

Cannot be true because csc(\theta)  is greater than zero in quadrant 2.

Step-by-step explanation:

When the csc of an angle is negative, since the cosecant function is defined as:

csc(\theta)=\frac{1}{sin(\theta)}

that means that the sin of the angle must be negative, and such cannot happen in the second quadrant. The sine function is positive in the first and second quadrant.

Therefore, the correct answer is:

Cannot be true because  csc(\theta)  is greater than zero in quadrant 2.

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Every possible cross section of a three dimensional figure is a circle what is that figure
Korolek [52]
A sphere. It is a perfectly circular and even shape, so all cross sections would be circles. 
5 0
2 years ago
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Prove that sinA-sin3A+sin5A-sin7A/cosA-cos3A-cos5A+cos7A= cot2A
MAVERICK [17]
Write the left side of the given expression as N/D, where
N = sinA - sin3A + sin5A - sin7A
D = cosA - cos3A - cos5A + cos7A
Therefore we want to show that N/D = cot2A.

We shall use these identities:
sin x - sin y = 2cos((x+y)/2)*sin((x-y)/2)
cos x - cos y = -2sin((x+y)/2)*sin((x-y)2)

N = -(sin7A - sinA) + sin5A - sin3A
    = -2cos4A*sin3A + 2cos4A*sinA
    = 2cos4A(sinA - sin3A)
    = 2cos4A*2cos(2A)sin(-A)
    = -4cos4A*cos2A*sinA

D = cos7A + cosA - (cos5A + cos3A)
   = 2cos4A*cos3A - 2cos4A*cosA
   = 2cos4A(cos3A - cosA)
   = 2cos4A*(-2)sin2A*sinA
   = -4cos4A*sin2A*sinA

Therefore
N/D = [-4cos4A*cos2A*sinA]/[-4cos4A*sin2A*sinA]
       = cos2A/sin2A
      = cot2A

This verifies the identity.
4 0
2 years ago
The epicenter of an earthquake is the point on Earth’s surface directly above the earthquake’s origin. A seismograph can be used
DaniilM [7]

Answer:

I used desmos and plotted all three circles. The point that hey all intersected was (-1, 3).

Step-by-step explanation:

5 0
2 years ago
If x = (10 − 3i) and y = (3 − 10i), then xy = -109i and x/y =
Yuliya22 [10]
Xy = -109i
We could find the value of i by substitute the algebraic form of x and y to the equation above

xy = -109i
(10 - 3i)(3 - 10i) = -109i
(10)(3) -3i(3) + 10(-10i) - 3i(-10i) = -109i
30 - 9i - 100i -30i² = -109i

multiply both side by -1
-30 + 9i + 100i + 30i² = 109i
30i² + 9i + 100i - 109i - 30 = 0
30i² - 30 = 0
30i² = 30
i² = 1
i = -1 or i = 1

Then find the value of x and y if i = -1
If i = -1, therefore
x = 10 - 3(-1)
x = 10 + 3
x = 13

y = (3 - 10i)
y = 3 - 10(-1)
y = 3 + 10
y = 13

x/y = 13/13 = 1

Then find the value of x and y if i = 1
x = 10 - 3(1)
x = 10 - 3
x = 7

y = (3 - 10i)
y = 3 - 10(1)
y = 3 - 10
y = -7

x/y = 7/-7 = -1

The value of x/y is either 1 or -1
7 0
2 years ago
Read 2 more answers
What is the constant of variation, k, of the direct variation, y = kx, through (5, 8)? k = – k equals negative StartFraction 8 O
Reil [10]

The value of constant of variation "k" is k = \frac{8}{5} \text{ or } 1.6

<em><u>Solution:</u></em>

Given that the direct variation is:

y = kx ----- eqn 1

Where "k" is the constant of variation

Given that the point is (5, 8)

<em><u>To find the value of "k" , substitute (x, y) = (5, 8) in eqn 1</u></em>

8 = k \times 5\\\\k = \frac{8}{5}\\\\k = 1.6

Thus the value of constant of variation "k" is k = \frac{8}{5} \text{ or } 1.6

3 0
2 years ago
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