Please someone help me

see explanation

Step-by-step explanation:

Using the identity

cos2Θ = 1 - 2sin²Θ, then

1 - 2sin²( - )

= cos [2( - )]

= sos( - Θ )

= coscosΘ + sin

= 0 × cosΘ + 1 × sinΘ

= 0 + sinΘ

= sinΘ = right side

 see proof below

Step-by-step explanation:

Use the Difference Identity: sin (A + B) = sin A cos B - cos A sin B

Use the following Half-Angle Identities:  

Use the Pythagorean Identity: cos²A + sin²A = 1   -->   sin²A = 1 - cos²A

Use the Unit Circle to evaluate:

Proof LHS → RHS

RHS = LHS:   sin A = sin A  

answer: the answer is 5/22

step-by-step explanation: if you subtract 17 from 22 the answer is 5 therefore the remainder of students are female

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.