Given WXYZ is a rhombus.

Prove WY bisects Angle ZWX and Angle XYZ.

ZX bisects Angle WZY and Angle YXW.

Diagonal of rhombus bisect the angles of it.

Step-by-step explanation:

Given: WXYZ is a rhombus.

To prove: WY bisects Angle ZWX and Angle XYZ.  ZX bisects Angle WZY and Angle YXW.

In rhombus WXYZ, WY is a diagonal.

In triangle WXY and triangle WZY,

       (sides of rhombus are equal)

       (sides of rhombus are equal)

      (Common side)

By SSS postulate,

      (CPCTC)

      (CPCTC)

Hence prove, that WY bisects Angle ZWX and Angle XYZ.

Similarly,

In rhombus WXYZ, ZX is a diagonal.

In triangle WXZ and triangle YXZ,

       (sides of rhombus are equal)

        (Common side)

       (sides of rhombus are equal)

By SSS postulate,

      (CPCTC)

      (CPCTC)

Hence prove, that ZX bisects Angle WZY and Angle YXW.

so firstly, we need to isolate the y variables to be able to solve these inequalities. firstly, add 0.5x on both sides of the first inequality and subtract 1.5x on both sides of the second inequality:

now since the slope is positive for the first inequality, this means that the line going upwards belongs to the first inequality, and the line going downwards belongs to the second inequality.

next, since y ≥ in the first inequality, this means that the shaded region will be above the first inequality's line, thus shading regions a and b.

next, since y ≤ in the second inequality, this means that the shaded region is going to be below this line, thus shading regions a and c.

since both lines shade region a, this means that region a is the solution.

step-by-step explanation:

mark brainliest!

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.