CL how do I avoid the “talk”

Mathematics

1 answer:

Aleksandr-060686 [28]3 years ago

8 0

Answer:

i don't know, start throwing up or something. can't talk while you're sick

You might be interested in

Find the exact value of tan (arcsin (two fifths)). For full credit, explain your reasoning.

Hitman42 [59]

$\bf sin^{-1}(some\ value)=\theta \impliedby \textit{this simply means}
\\\\\\
sin(\theta )=some\ value\qquad \textit{now, also bear in mind that}
\\\\\\
sin(\theta)=\cfrac{opposite}{hypotenuse}\qquad 
\qquad 
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}\\\\
-------------------------------\\\\$

$\bf sin^{-1}\left( \frac{2}{5} \right)=\theta \impliedby \textit{this simply means that}
\\\\\\
sin(\theta )=\cfrac{2}{5}\cfrac{\leftarrow opposite}{\leftarrow hypotenuse}\qquad \textit{now let's find the adjacent side}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}$

$\bf \pm \sqrt{5^2-2^2}=a\implies \pm\sqrt{21}=a
\\\\\\
\textit{we don't know if it's +/-, so we'll assume is the + one}\quad \sqrt{21}=a\\\\
-------------------------------\\\\
tan(\theta)=\cfrac{opposite}{adjacent}\qquad \qquad tan(\theta)=\cfrac{2}{\sqrt{21}}
\\\\\\
\textit{and now, let's rationalize the denominator}
\\\\\\
\cfrac{2}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies \cfrac{2\sqrt{21}}{(\sqrt{21})^2}\implies \cfrac{2\sqrt{21}}{21}
$

7 0

3 years ago

Your family rented a car for a trip the car rental cost $35 per day plus $.30 per mile after a one day rental The bill was $74 h

loris [4]

Answer:

120 miles

Step-by-step explanation:

You payed $74, 35 of which came from the daily charge, meaning $36 came from the total miles driven. 36 / 0.30 = 120

4 0

3 years ago

The rate of change of the mass, A, of salt at time t is proportional to the square of the mass of salt present at time t. Initia

Brums [2.3K]

Answer:

(a) -dA/dt = kA², A₀ = 10

(b) A =10/(1+ kt)

Step-by-step explanation:

(a) Find the IVP

A differential equation with an initial condition y₀ = f(x₀) is called an initial value problem.

The rate of decrease of A is proportional to A², and A₀ = 10, so the IVP is

-dA/dt = kA², A₀ = 10

(b) Solve the IVP

$\begin{array}{rcl}-\dfrac{\text{d}A}{\text{d}t} & = & kA^{2}\\\\\dfrac{\text{d}A}{A^{2}} & =&-k\text{d}t\\\\-\dfrac{1}{A} & = & -kt + C\\\\\dfrac{1}{A} & = & kt + C\\\\\end{array}$

Apply the initial condition: A₀ = 10 (when t = 0)

$\dfrac{1}{10} = C\\\\\dfrac{1}{A} = kt + \dfrac{1}{10}\\\\\dfrac{10}{A} = 10kt + 1\\\\\mathbf{A} = \mathbf{\dfrac{10}{1+ kt}}$

(c) Find the time when A(t) < 1

(i) Find the value of k (A₁₀ = 4)

$\begin{array}{rcl}A& =&\dfrac{10}{1+ kt}\\\\4 & =& \dfrac{10}{1 + 10k}\\\\4 + 40 k & = & 10\\\\40k & = & 6\\k & = & 0.15\\\end{array}\\\\\mathbf{A} =\mathbf{ \dfrac{10}{0.15t + 1}}$

(ii) Find t when A < 1

$\begin{array}{rcl}A(t) & < & 1\\\dfrac{10}{0.15t + 1} & < & 1\\\\10 & < & 0.15t + 1\\9 & < & 0.15t\\t & > & 60\\\\\end{array}\\\mathbf{A < 1} \textbf{ when }\mathbf{ t >60}$