I think the correct answer would be D. The two parallelograms are congruent because a rotation does not change size and shape.
Let First Sphere be the Original Sphere
its Radius be : r
We know that Surface Area of the Sphere is : 4π × (radius)²
⇒ Surface Area of the Original Sphere = 4πr²
Given : The Radius of Original Sphere is Doubled
Let the Sphere whose Radius is Doubled be New Sphere
⇒ Surface of the New Sphere = 4π × (2r)² = 4π × 4 × r²
But we know that : 4πr² is the Surface Area of Original Sphere
⇒ Surface of the New Sphere = 4 × Original Sphere
⇒ If the Radius the Sphere is Doubled, the Surface Area would be enlarged by factor : 4
Answer:
x=3
Step-by-step explanation:
4-2x-4=-2-4
-2x=-6
\frac{-2x}{-2}=\frac{-6}{-2}
x=3
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