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2 years ago

If ³√a=am , what is m

Mathematics

1 answer:

Zina [86]2 years ago

8 0

$\sqrt[3]{a} = {a}^{m} \\ = > { a}^{ \frac{1}{3} } = {a}^{m} \\ = > m = \frac{1}{3}$

Answer:

$m = \frac{1}{3}$

Hope you could understand.

If you have any query, feel free to ask.

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nikdorinn [45]

Answer:

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Step-by-step explanation:

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3 years ago

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A broken faucet leaks 1 gallon of water every 1 and1/3 months the amount of months that pass m varies directly with the amount o

Simora [160]

$\bf \qquad \qquad \textit{direct proportional variation}\\\\
\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------\\\\$

$\bf \textit{\underline{m} varies directly with the amount of gallons}\implies \stackrel{months}{m}=k\stackrel{gallons}{g}
\\\\\\
\textit{we also know that }
\begin{cases}
m=1\frac{1}{3}\\
\qquad \frac{4}{3}\\
g=1
\end{cases}\implies \cfrac{4}{3}=k1\implies \cfrac{4}{3}=k
\\\\\\
\boxed{m=\cfrac{4}{3}g}\\\\
-------------------------------\\\\
\textit{what's \underline{m} when g=7?}\qquad m=\cfrac{4}{3}\cdot 7\implies m=\cfrac{28}{3}\implies m=9\frac{1}{3}$

6 0

3 years ago

A solid right pyramid has a square base. The length of the base edge is4 cm and the height of the pyramid is 3 cm period what is

Illusion [34]

Answer:

The volume of this pyramid is 16 cm³.

Step-by-step explanation:

The volume $V$ of a solid pyramid can be given as:

$\displaystyle V = \frac{1}{3} \cdot b \cdot h$ ,

where

$b$ is the area of the base of the pyramid, and
$h$ is the height of the pyramid.

Here's how to solve this problem with calculus without using the previous formula.

Imaging cutting the square-base pyramid in half, horizontally. Each horizontal cross-section will be a square. The lengths of these squares' sides range from 0 cm to 3 cm. This length will be also be proportional to the vertical distance from the vertice of the pyramid.

Refer to the sketch attached. Let the vertical distance from the vertice be $x$ cm.

At the vertice of this pyramid, $x = 0$ and the length of a side of the square is also $0$ .
At the base of this pyramid, $x = 3$ and the length of a side of the square is $4$ cm.

As a result, the length of a side of the square will be

$\displaystyle \frac{x}{3}\times 4 = \frac{4}{3}x$ .

The area of the square will be

$\displaystyle \left(\frac{4}{3}x\right)^{2} = \frac{16}{9}x^{2}$ .

Integrate the area of the horizontal cross-section with respect to $x$

from the top of the pyramid, where $x = 0$ ,
to the base, where $x = 3$ .

$\displaystyle \begin{aligned}\int_{0}^{3}{\frac{16}{9}x^{2}\cdot dx} &= \frac{16}{9}\int_{0}^{3}{x^{2}\cdot dx}\\ &= \frac{16}{9}\cdot \left(\frac{1}{3}\int_{0}^{3}{3x^{2}\cdot dx}\right) & \text{Set up the integrand for power rule}\\ &= \left.\frac{16}{9}\times \frac{1}{3}\cdot x^{3}\right|^{3}_{0}\\ &= \frac{16}{27}\times 3^{3} \\ &= 16\end{aligned}$ .