The diagonals of a rhombus are perpendicular bisectors of each other. You can use the Pythagorean theorem. If the diagonals are length "a" and "b", the side length of the rhombus (s) is
s = (1/2)√(a²+b²)
The answer would be (x+3)^2+4
I suppose you just have to simplify this expression.
(2ˣ⁺² - 2ˣ⁺³) / (2ˣ⁺¹ - 2ˣ⁺²)
Divide through every term by the lowest power of 2, which would be <em>x</em> + 1 :
… = (2ˣ⁺²/2ˣ⁺¹ - 2ˣ⁺³/2ˣ⁺¹) / (2ˣ⁺¹/2ˣ⁺¹ - 2ˣ⁺²/2ˣ⁺¹)
Recall that <em>n</em>ª / <em>n</em>ᵇ = <em>n</em>ª⁻ᵇ, so that we have
… = (2⁽ˣ⁺²⁾ ⁻ ⁽ˣ⁺¹⁾ - 2⁽ˣ⁺³⁾ ⁻ ⁽ˣ⁺¹⁾) / (2⁽ˣ⁺¹⁾ ⁻ ⁽ˣ⁺¹⁾ - 2⁽ˣ⁺²⁾ ⁻ ⁽ˣ⁺¹⁾)
… = (2¹ - 2²) / (2⁰ - 2¹)
… = (2 - 4) / (1 - 2)
… = (-2) / (-1)
… = 2
Another way to get the same result: rewrite every term as a multiple of <em>y</em> = 2ˣ :
… = (2²×2ˣ - 2³×2ˣ) / (2×2ˣ - 2²×2ˣ)
… = (4×2ˣ - 8×2ˣ) / (2×2ˣ - 4×2ˣ)
… = (4<em>y</em> - 8<em>y</em>) / (2<em>y</em> - 4<em>y</em>)
… = (-4<em>y</em>) / (-2<em>y</em>)
… = 2
Solve for x
Divide both sides by -4
X= - 10/4
Now simplify by dividing top and bottom by 2
X= -5/2 or - 2 1/2
Hey There
According to me,
no 1 is
SQUARE
no. 2 is
CONGRUENT
HOPE THIS HELPS
