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Airida [17]
2 years ago
8

Please solve this quick.

Mathematics
1 answer:
Ivenika [448]2 years ago
6 0

Should be JK

If not sorry

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PLEASE HELP!!
erik [133]
2.=c.\\\\2\sin4x\cos4x=2\sin(2\cdot4x)=2\sin8x\\\\Used:\\\sin2\alpha=2\sin\alpha\cos\alpha

1.=b.\\\\\csc x-\sin x=\dfrac{1}{\sin x}-\dfrac{\sin^2x}{\sin x}=\dfrac{1-\sin^2x}{\sin x}=\dfrac{\cos^2x}{\sin x}\\\\=\dfrac{\cos x\cos x}{\sin x}=\cos x\cdot\dfrac{\cos x}{\sin x}=\cos x\cot x\\\\Used:\\\csc x=\dfrac{1}{\sin x}\\\\\sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\\\\\cot x=\dfrac{\cos x}{\sin x}

3.=a.\\\\\dfrac{\sin x-1}{\sin x+1}=\dfrac{\sin x-1}{\sin x+1}\cdot\dfrac{\sin x+1}{\sin x+1}=\dfrac{\sin^2x-1^2}{(\sin x+1)^2}=\dfrac{\sin^2x-1}{(\sin x+1)^2}\\\\=\dfrac{-(1-\sin^2x)}{(\sin x+1)^2}=\dfrac{-\cos^2x}{(\sin x+1)^2}\\\\Used:\\(a-b)(a+b)=a^2-b^2\\\\\sin^2x+\cos^2x=1\to \cos^2x=1-\sin^2x
8 0
3 years ago
A tower is 105' tall. it casts a shadow 33' long. what is the angle of depression?
KiRa [710]
Tan(x)= 33/105

x = arctan(33/105)

x = 17.44718842 degrees.
7 0
2 years ago
The diagram below shows 5 identical bowls stacked one inside the other. The height of 1 bowl is 2 inches. The height of 5 bowls
padilas [110]

(Bowls, Height) (1, 2)  (5,5)

Slope is (5-2)/(5-1) = 3/4 inch
y = (3/4)x + b
(2) = (3/4)(1) + b
(2)-(3/4) = b

B=1.25.  Y= 0.75*x + 1.25. 

Part B
X is the number of bowls in the stack and Y is the corresponding height of the stack.

 

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=f%28x%29%20%3D%20%28x%20-%20%20%5Cfrac%7B2%7D%7B9%7D%20%29%28x%20%2B%20%20%5Cfrac%7B1%7D%7B2%7
Aleonysh [2.5K]
f(x) = (x -  \dfrac{2}{9} )(x +  \dfrac{1}{2} )

\text {When } f(x) = 0 :

(x -  \dfrac{2}{9} )(x +  \dfrac{1}{2} ) = 0

(x -  \dfrac{2}{9} ) = 0  \text { or }  (x +  \dfrac{1}{2} ) = 0

x =  \dfrac{2}{9}  \text { or } x =  -\dfrac{1}{2}

3 0
3 years ago
Find the value of all trigonometric functions of 135°
Vera_Pavlovna [14]

Answer:

sin (- 135°)= – sin 135°= – sin (1 × 90°+ 45°) = – cos 45° = – 1√2

cos (- 135°)= cos 135°= cos (1 × 90°+ 45°) = – sin 45°= – 1√2

tan (- 135°) = – tan 135° = – tan ( 1 × 90° + 45°) = – (- cot 45°) = 1

csc (- 135°)= – csc 135°= – csc (1 × 90°+ 45°)= – sec 45° = – √2

sec (- 135°)= sec 135°= sec (1 × 90°+ 45°)= – csc 45°= – √2

cot (- 135°) = – cot 135° = – cot ( 1 × 90° + 45°) = – (-tan 45°) = 1

Step-by-step explanation:

hope this helps

5 0
2 years ago
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