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sasho [114]
3 years ago
14

4 less than the quantity of 8 times n

Mathematics
1 answer:
kirza4 [7]3 years ago
6 0
8n-4
____________________
:)
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What is the correct order for solving any literal multi-variable equation
KatRina [158]

Answer:

Step-by-step explanation:

Umm well  mean you just have to use like terms. start by doing any math until there are no parenthesis, and then you basically just use like terms to solve it.  (like if you have (a^{2} b)(ab) then you use like terms) I would need specifics to be able to evaluate more, but I hope this helps!

5 0
3 years ago
Show why the limit as x approaches 0 (csc(x)-cot(x)) involves an indeterminate form, and then prove that the limit equals 0.
Nadusha1986 [10]

Answer with Step-by-step explanation:

We are given that \lim_{x\rightarrow 0 }(csc(x)-cot(x))

We have to prove that why the limit x approaches 0(csc(x)-cot(x)) involves an indeterminate form and prove that the limit equals to 0.

\lim_{x\rightarrow 0 }(\frac{1}{sinx}-\frac{cosx}{sinx})

Because csc(x)=\frac{1}{sinx} andcot(x)=\frac{cosx}{sinx}

\lim_{x\rightarrow 0 }(\frac{1-cosx}{sinx})

\frac{1-cos0}{sin0}

We know that cos 0=1 and sin 0=0

Substitute the values then we get

\frac{1-1}{0}=\frac{0}{0}

We know that \frac{0}{0} is indeterminate form

Hence, the limit x approaches 0(csc(x)-cot(x)) involves an indeterminate form.

L'hospital rule:Apply this rule and  differentiate numerator and denominator separately when after applying \lim_{x\rightarrow a }we get indeterminate form\frac{0}{0}

Now,using L' hospital rule

\lim_{x\rightarrow 0 }\frac{0+sinx}{cosx}

because \frac{dsinx}{dx}=cosx,\frac{dcosx}{dx}=-sinx}

Now, we get

\lim_{x\rightarrow 0 }\frac{sinx}{cos x}

\frac{sin0}{cos0}

\frac{0}{1}=0

Hence,\lim_{x\rightarrow 0 }(csc(x)-cot(x))=0

5 0
2 years ago
Sin theata=3/5, then cos theata= ?
Elena-2011 [213]

\bf sin(\theta )=\cfrac{\stackrel{opposite}{3}}{\stackrel{hypotenuse}{5}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}
\\\\
c^2=a^2+b^2\implies \sqrt{c^2-b^2}=a
\qquad
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\sqrt{5^2-3^2}=a\implies \pm\sqrt{25-9}=a\implies \pm 4=a
\\\\[-0.35em]
\rule{34em}{0.25pt}\\\\
~\hfill cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 4}}{\stackrel{hypotenuse}{5}}~\hfill

5 0
3 years ago
Which is the equation of the line that passes through the point (-5, -2)
Lilit [14]

Answer:

F) y=3x+13

Step-by-step explanation:

y-y1=m(x-x1)

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y=3x+15-2

y=3x+13

5 0
2 years ago
During one month there were 7 days of precipitation.What if there had only been 3 days of precipitation that month?How would tha
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3 years ago
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