We are given with an isosceles triangle having a vertex on the curve given y =<span>27-x^2</span> .
The area of the triangle, A= xy = x (27-x^2)
A' = 27-x^2-2x^2 = 0
x = 3
Amax = 3(27-9) = 54 units2
Answer: (14 , 15/2)
Step-by-step explanation:
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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
Consider
.. X = {1, 2, 3, 4}, x = 4
.. Y = {2, 3, 4, 5, 6}, y = 5, w = 3
The elements in X or Y (X ∪ Y) are {1, 2, 3, 4, 5, 6}, n = 6.
.. n = 6 = 4 + 5 - 3
Note that if we just add x and y, we count the common elements twice. In order to just count the common elements once, we need to subtract that count from the total of x and y.
selection B is appropriate.
Answer:
You need to have both the opposite and adjacent side of the right angled triangle. Tan = Opp/Adj
Step-by-step explanation:
Tangents are pretty much everywhere. A good example is maybe a slide at the playground. You’d need the tangent to figure out the angle of the slide and how it is.
The radius from the center of a circle to the point of tangency point shows that it would be perpendicular to the tangent line considering that anything with a radius is circular and the tangent line is… a line, making it impossible to be parallel.
