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Tpy6a [65]
3 years ago
7

What is the vertical component of a 100 pound force applied at 30° above the horizontal?

Mathematics
1 answer:
lutik1710 [3]3 years ago
4 0
That would be 100 * sin 30 = 100 * 0.5 = 50 pounds
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Evaluate : limx→ tan2-sin2x x3
GrogVix [38]

Given:

\lim _{x\to0}\frac{\tan 2x-\sin 2x}{x^3}

Solve:

\lim _{x\to0}\frac{\tan 2x-\sin 2x}{x^3}

Use l'hopital's rule:

\begin{gathered} =\lim _{x\to0}\frac{\frac{d}{dx}(-\sin 2x+\tan 2x)}{\frac{d}{dx}(x^3)} \\ =\lim _{x\to0}\frac{-2\cos (2x)+2\tan ^2(2x)+2}{3x^2} \end{gathered}

Simplify:

\begin{gathered} =\lim _{x\to0}\frac{-2\cos (2x)+2\tan ^2(2x)+2}{3x^2} \\ =\lim _{x\to0}\frac{2(-\cos (2x)+\tan ^2(2x)+1)}{3x^2} \end{gathered}

Apply the constant multiple rule:

\begin{gathered} \lim _{x\to0}cf(x)=c\lim _{x\to0}f(x) \\ \text{With c=}\frac{2}{3} \\ f(x)=\frac{-\cos (2x)+\tan ^2(2x)+1}{x^2} \end{gathered}\begin{gathered} =\frac{2\lim _{x\to0}\frac{-\cos (2x)+\tan ^2(2x)+1}{x^2}}{3} \\ =\frac{2\lim _{x\rightarrow0}\frac{(4\tan ^2(2x)+4)\tan (2x)+2\sin (2x)}{2x}}{3} \end{gathered}

Similary :

\begin{gathered} =\frac{2\lim _{x\to0}(2\cos (2x)+12\tan ^4(2x)+16\tan ^2(2x)+4)}{3} \\ =\frac{2(6)}{3} \\ =4 \end{gathered}

8 0
1 year ago
Every morning, halle goes to school with a 1-liter bottle of water. she drinks 1/4 of the bottle before school starts and 2/3 of
Nuetrik [128]
I think it's 1/6. Because before school she drinks 1/4 then before lunch she drink 2/3 if 1/4 so you would multiply. 1*2=2. 4*3=12. So you get 2/12 or if you simplify, 1/6
8 0
3 years ago
Picture attached multiple choice :3
nikklg [1K]

it is C. i think because this i looked it up trust me

8 0
3 years ago
In a class of 40 students, 15 offer physics, 20 offer history and 3 offer both physics and history. How many student offer neith
Ainat [17]

Answer:

2 students.

Step-by-step explanation:

add together all of the students and then subtract that from 40.

20+15+3 = 38

40-38 = 2

5 0
3 years ago
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I need help now I’m stuck!
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Whats the question?
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2 years ago
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