How do I find the surface area of a triangle?

1 answer:

8 0

It depends on what you know about the triangle.

The usual formula taught in Geometry class is
.. A = (1/2)*b*h
where b is the length of the base, and h is the height (altitude)

If you know two sides (a and b) and the angle between them,
.. A = (1/2)*a*b*sin(θ)

If you know three sides (a, b, c) then Heron's formula applies:
.. s = (a+b+c)/2
.. A = √(s*(s -a)*(s -b)*(s -c))

If you know two sides and an angle not between them, then you can use the Law of Sines to find the angle between them and use the formula above. In some of these cases, the triangle may be incompletely specified, so there may be two different legitimate values for the area.

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Y=sqrt(x)(8x-5) find the derivative

garri49 [273]

Answer:

$\displaystyle y' = \frac{24x - 5}{2\sqrt{x}}$

General Formulas and Concepts:

Algebra I

Exponentials [Fractions] - Are radicals
Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Product Rule: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Step-by-step explanation:

Step 1: Define

$\displaystyle y = \sqrt{x}(8x - 5)$

Step 2: Differentiate

$\displaystyle f(x) = \sqrt{x}, \ g(x) = (8x - 5)$

Product Rule: $\displaystyle y' = \frac{d}{dx}[\sqrt{x}] \cdot (8x - 5) + \sqrt{x} \cdot \frac{d}{dx}[(8x - 5)]$
Rewrite: $\displaystyle y' = \frac{d}{dx}[x^{\frac{1}{2}}] \cdot (8x - 5) + \sqrt{x} \cdot \frac{d}{dx}[(8x - 5)]$
Basic Power Rule: $\displaystyle y' = \frac{1}{2}x^{\frac{1}{2} - 1} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8x^{1 - 1}$
Simplify: $\displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8x^{0}$
Rewrite: $\displaystyle y' = \frac{1}{2x^{\frac{1}{2}}} \cdot (8x - 5) + \sqrt{x} \cdot 1 \cdot 8$
Multiply: $\displaystyle y' = \frac{8x + 5}{2x^{\frac{1}{2}}} + 8\sqrt{x}$
Rewrite: $\displaystyle y' = \frac{8x + 5}{2\sqrt{x}} + 8\sqrt{x}$
Add/Rewrite: $\displaystyle y' = \frac{24x - 5}{2\sqrt{x}}$