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3 years ago

(05.01)Which equation can be used to calculate the area of the shaded triangle in the figure below?

Mathematics

1 answer:

Marina86 [1]3 years ago

7 0

Equation of shaded triangle is $\frac{1}{2} (14\times8)$ .

Solution:

Length of the rectangle = 14 ft

Breadth of the rectangle = 8 ft

Area of the rectangle = length × breadth

= 14 × 8

In this rectangle the diagonal splits the rectangle into two equal halves.

Area of the shaded triangle = Half of area of rectangle

$$=\frac{1}{2} (14\times8)$

Area of the shaded triangle = $\frac{1}{2} (14\times8)$

Hence the equation can be used to calculate the area of the shaded

triangle is $\frac{1}{2} (14\times8)$ .

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If we know this, the variables for this problem are assuming the function is 10(5.5)^x:

x2: 2x1: 1f(x2): 10(5.5)^(2) = 302.5f(x1): 10(5.5)^(1)= 55

This means:y = 302.5 - 55 = 247.5x = 2 - 1 = 1

Remember: the equation for avg rate of change is y/x

So, our average rate of change for the function on the interval [1,2] is 247.5 (y/x = 247.5/1)

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3 years ago

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Derivative, by first principle<br><img src="https://tex.z-dn.net/?f=%20%5Ctan%28%20%5Csqrt%7Bx%20%7D%20%29%20" id="TexFormula1"

vampirchik [111]

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}h$

Employ a standard trick used in proving the chain rule:

$\dfrac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\cdot\dfrac{\sqrt{x+h}-\sqrt x}h$

The limit of a product is the product of limits, i.e. we can write

$\displaystyle\left(\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan x}{\sqrt{x+h}-\sqrt x}\right)\cdot\left(\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}h\right)$

The rightmost limit is an exercise in differentiating $\sqrt x$ using the definition, which you probably already know is $\dfrac1{2\sqrt x}$ .

For the leftmost limit, we make a substitution $y=\sqrt x$ . Now, if we make a slight change to $x$ by adding a small number $h$ , this propagates a similar small change in $y$ that we'll call $h'$ , so that we can set $y+h'=\sqrt{x+h}$ . Then as $h\to0$ , we see that it's also the case that $h'\to0$ (since we fix $y=\sqrt x$ ). So we can write the remaining limit as

$\displaystyle\lim_{h\to0}\frac{\tan\sqrt{x+h}-\tan\sqrt x}{\sqrt{x+h}-\sqrt x}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{y+h'-y}=\lim_{h'\to0}\frac{\tan(y+h')-\tan y}{h'}$

which in turn is the derivative of $\tan y$ , another limit you probably already know how to compute. We'd end up with $\sec^2y$ , or $\sec^2\sqrt x$ .

So we find that

$\dfrac{\mathrm d\tan\sqrt x}{\mathrm dx}=\dfrac{\sec^2\sqrt x}{2\sqrt x}$

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