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AnnyKZ [126]
3 years ago
9

How much water can this container hold? use 3.14 to apporximate pi.

Mathematics
1 answer:
zimovet [89]3 years ago
4 0

Answer:well you need to know the actual height and diameter of the container if it's a curved shape

Step-by-step explanation:

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What is the inverse of f(x)=2x^3
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Answer:

f^-1 (x)=(x/2)^(1/3)

Step-by-step explanation:

y=2x^3

x=2y^3

2y^3=x

y^3=x/2

y=(x/2)^(1/3)

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Help please with this
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Rewrite 9cos 4x in terms of cos x.
rosijanka [135]
\bf \qquad \textit{Quad identities}\\\\
sin(4\theta )=
\begin{cases}
8sin(\theta )cos^3(\theta )-4sin(\theta )cos(\theta )\\
4sin(\theta )cos(\theta )-8sin^3(\theta )cos(\theta )
\end{cases}
\\\\\\
cos(4\theta)=8cos^4(\theta )-8cos^2(\theta )+1\\\\
-------------------------------\\\\
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\\\\\\
72cos^4(x)-72cos^2(x)+9


---------------------------------------------------------------------------

as far as the previous one on the 2tan(3x)

\bf tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\qquad tan({{ \alpha}} + {{ \beta}}) = \cfrac{tan({{ \alpha}})+ tan({{ \beta}})}{1- tan({{ \alpha}})tan({{ \beta}})}\\\\
-------------------------------\\\\

\bf 2tan(3x)\implies 2tan(2x+x)\implies 2\left[  \cfrac{tan(2x)+tan(x)}{1-tan(2x)tan(x)}\right]
\\\\\\
2\left[  \cfrac{\frac{2tan(x)}{1-tan^2(x)}+tan(x)}{1-\frac{2tan(x)}{1-tan^2(x)}tan(x)}\right]\implies 2\left[ \cfrac{\frac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}}{\frac{1-tan(x)-2tan^3(x)}{1-tan^2(x)}} \right]
\\\\\\

\bf 2\left[ \cfrac{2tan(x)+tan(x)-tan^3(x)}{1-tan^2(x)}\cdot \cfrac{1-tan^2(x)}{1-tan(x)-2tan^3(x)} \right]
\\\\\\
2\left[ \cfrac{3tan(x)-tan^3(x)}{1-tan^2(x)-2tan^3(x)} \right]\implies \cfrac{6tan(x)-2tan^3(x)}{1-tan^2(x)-2tan^3(x)}
4 0
3 years ago
a number is divided by 3 and the result is added to 1. the new result is doubled to give the final result 80. whats the number
Gnesinka [82]
78, because you have to do the inverse operation so 80/2= 40 then 40-1= 39 and 39*3 = 78
6 0
3 years ago
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