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Strike441 [17]
3 years ago
14

Which of the following inverse functions are defined for x=-1/2? Select all thst apply.

Mathematics
2 answers:
daser333 [38]3 years ago
4 0
All but the last one I think but don't quote me
ludmilkaskok [199]3 years ago
3 0

Solution:

Function      Domain

sin−1(x)            [−1,1]   →→→→→Defined

cos−1(x)           [−1,1]  →→→→→Defined

tan−1(x)          (−∞,∞)  →→→→→Defined

cot−1(x)          (−∞,∞)  →→→→→Defined

sec−1(x)        (−∞,−1]∪[1,∞)  →→→→→ not Defined

cosec−1(x)     (−∞,−1]∪[1,∞)  →→→→→not Defined

As, you can see that, all inverse trigonometric functions are defined for, x=\frac{-1}{2}, except sec−1(x)   and cosec−1(x)

, whose domain is not valid or do not exist when , x=\frac{-1}{2}.

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Answer:

A. 9.8 ft

Step-by-step explanation:

first convert the degrees to radians:

80° x (pi/180) ≈ 1.396

arc length = r x θ (always in radians)

= 7 x 1.396

= 9.772 ≈ 9.8 ft

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There are 205 total students in a school. 20% of the students are 5th
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41 students

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2 years ago
Calc help pretty please
Gre4nikov [31]

Answer:

A.  77.4

Step-by-step explanation:

<u>Linear approximation formula</u>

L(x)=f(a)+f'(a)(x-a)

Given function:

f(x)=(x+1)^2

\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If  $y=f(u)$  and  $u=g(x)$  then:\\\\$\dfrac{\text{d}y}{\text{d}x}=\dfrac{\text{d}y}{\text{d}u}\times \dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}

\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If  $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=xn^{n-1}$\\\end{minipage}}

\boxed{\begin{minipage}{4 cm}\underline{Differentiating $ax$}\\\\If  $y=ax$, then $\dfrac{\text{d}y}{\text{d}x}=a$\\\end{minipage}}

Use the chain rule to differentiate the function.

\textsf{Let }\:y=u^2\:\textsf{ where }u=(x+1)

Differentiate the two parts separately:

  y=u^2 \implies \dfrac{\text{d}y}{\text{d}u}=2u

  u=x+1 \implies \dfrac{\text{d}u}{\text{d}x}=1

Put everything back into the chain rule formula:

\begin{aligned} \implies \dfrac{\text{d}y}{\text{d}x} & =2u \times 1\\ & = 2u \\ & = 2(x+1)\\ & = 2x+2 \end{aligned}

\textsf{Therefore, }\:f'(x)=2x+2..

The <u>linear approximation</u> at a = 8 is:

\begin{aligned}L(x) & =f(a)+f'(a)(x-a)\\\\\implies L(x) & = f(8)+f'(8)(x-8)\\& = (8+1)^2+(2(8)+2)(x-8)\\& = 81+18(x-8)\\& = 18x-63\end{aligned}

Finally, substitute x = 7.8 into the <u>linear approximation equation</u>:

\begin{aligned}\implies L(7.8) & =18(7.8)-63\\& = 140.4-63\\& = 77.4\end{aligned}

4 0
1 year ago
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