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777dan777 [17]
3 years ago
10

Let θ be an angle in quadrant III such csc(θ)=-(13)/(5). . Find the exact values of tan θ and cosθ ?

Mathematics
1 answer:
LenaWriter [7]3 years ago
3 0
Given that the angle is in quadrant II such that <span>csc(θ) = -(13)/(5), the expression can be expressed into 1/sin </span>(θ) = -(13)/(5). θ can be calculated through the calculator equal to -22.62 degrees. Since the angle is in quadrant II, <span>θ is equal to 180-22.62 ot 157.38 degrees. Tan </span>θ then is equal to -5/12, cos <span>θ is equal to -12/13.</span>
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Can u help me with this
daser333 [38]

Answer:

acc to ques....perimeter of figure= 3a^2+15a+9

let missing dimension be x

= x+10a-2+a^2-7+5a=3a^2+15a+9

x+15a+a^2-9=3a^2+15a+9

x=2a^2+18

6 0
3 years ago
Find the circumference of the circle below to the nearest tenth Use 3.14 for a
AnnyKZ [126]

Answer:

125.6

Step-by-step explanation:

all you have to do is multiple the 3.14 with 40 cm to get 125.6 cm

8 0
2 years ago
Please help and explain would I still set this up with law of sine or use cosine
Svetach [21]

You would use the tangent first, because the opposite of 40 would be x and the adjacent would be 30ft

4 0
3 years ago
Which equation will be parallel to y=6 A)Y= -6x B)X=6 C)y=1 D)Y=6x
galben [10]

Answer:

A

Step-by-step explanation:

4 0
3 years ago
Find the major axis for the ellipse <br> x² + 16y2-96y + 128 = 0
Softa [21]

The major axis for the ellipse, x² + 16y² - 96y + 128 = 0 is the x-axis

To answer the question, we need to write it in the standard form of the equation of an ellipse

<h3>Equation of an ellipse</h3>

The equation of an ellipse centered at  (h,k) is

(x - h)²/a² + (y - k)²/b² (1) where a > b and the major axis is parallel to the x axis

Given x² + 16y² - 96y + 128 = 0, we convert it into the standard equation of an ellipse.

So, x² + 16y² - 96y + 128 = 0

Dividing through by 16, we have

x²/16 + 16y²/16  - 96y/16 + 128/16 = 0/16

x²/16 + y² - 6y + 8 = 0

Completing the square in y by adding and subtracting (-6/2)² = (-3)²

x²/16 + y² - 6y + (-3)² - (-3)² + 8 = 0

x²/16 + (y - 3)² - 9 + 8 = 0

x²/16 + (y - 3)² - 1 = 0

x²/16 + (y - 3)² = 1

x²/4² + (y - 3)²/1² = 1  (2)

Comparing equations (1) and (2), we have that a = 4 and b = 1.

Since a = 4 > b = 1, the major axis for the ellipse is the x-axis

So, the major axis for the ellipse, x² + 16y² - 96y + 128 = 0 is the x-axis

Learn more about ellipse here:

brainly.com/question/26679189

#SPJ1

8 0
2 years ago
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