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3 years ago

Find d / dx ( e^in(x)).

1 answer:

5 0

let's first off apply a log rule of cancellation, keeping in mind that, first off is ln(), not in(), and that ln() is just a shortcut to logₑ.

$\bf \textit{Logarithm Cancellation Rules}
\\\\
log_a a^x = x\qquad \qquad \stackrel{\stackrel{\textit{we'll use this one}}{\downarrow }}{a^{log_a x}=x}
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
e^{ln(x)}\implies e^{log_e(x)}\implies x
\\\\\\
\cfrac{d}{dx}\left[ e^{ln(x)} \right]\implies \cfrac{d}{dx}[x]\implies 1$

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Thats the question

elixir [45]

Answer:

160 in ^2

Step-by-step explanation:

First we need to find the area of each shape and add them together

1. Triangles:

The formula for a triangle is (base x height)/2, so we can replace them as (5 * 4)/2 * 2( Because there are two triangles), so therefore the two triangles will add up to 20 inches

2. The Rectangles

The Big Rectangle:

The big rectangle is just l x w or 5 * 20 which is 100

The small rectangle:

To find the width of the small rectangle you have to do 20 - (5 + 5) because we are not including the triangles. 20 - (5 + 5) = 10, so that would be 10 * 4 = 40.

3.Add them together

20 + 100 + 40 = 160 inches ^2

Hope this helps!!!

4 0

2 years ago

$\triangle abc$ and $\triangle dbc$ share $bc$. $ab = 5\ \text{cm}$, $ac = 12\ \text{cm}$, $dc = 8\ \text{cm}$, and $bd = 20\ \t

Ratling [72]

What terms govern the length of this side?The basic rule of the triangle.First side length must be less than the sum of the other two sides.So to find X we must take the largest side of the triangles and compare them with amounts from other sides.

5+x>12
8+x>20 and it's system

x>7
x>12
general solution is x>12
The least possible integral is 13.

PS: It's may be yet 12, but in this case, triangle BCD become segment.

6 0

3 years ago

- Yesterday Billy earned $30 trimming

horsena [70]

10x8=80 dollars that day
80+30=$110

Hope this helps!! Have a good day

5 0

3 years ago

Would appreciate the help !

aleksandr82 [10.1K]

This is one pathway to prove the identity.

Part 1

$\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 2

$\frac{\sin^2(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)-\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-(\cos(\theta)-\cos^2(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{\sin^2(\theta)-\cos(\theta)+\cos^2(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 3

$\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\$

As the steps above show, the goal is to get both sides be the same identical expression. You should only work with one side to transform it into the other. In this case, the left side transforms while the right side stays fixed the entire time. The general rule is that you should convert the more complicated expression into a simpler form.

We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity $\sin^2(\theta)+\cos^2(\theta) = 1$ in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

3 0

2 years ago

Light travels at a speed of 3.0 ´ 108 m/s. If it takes light from the sun 5.0 ´ 102 s to reach Earth, what is the distance betwe

galben [10]

(5.0 x 10²) x (3.0 x 10⁸) = 1.5 x 10¹¹ meters = 1.5 x 10⁸ km.

5 0

3 years ago