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3 years ago

Convert: 9 cubic meters to cubic feet (to the nearest tenth).

Mathematics

2 answers:

chubhunter [2.5K]3 years ago

7 0

Answer C. 317.8 would be your answer:) Hope this helps!

lions [1.4K]3 years ago

5 0

9 m^3 * ( (1 ft)^3 )/( (0.3048 m)^3 ) = 317.8 ft^3

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Consider the radical equation √n+ 4 = n-2.

muminat

Answer:

Pretty sure neither of them work. So prolly the last one. Hope I’m right :)

3 0

3 years ago

Solve in simplest form<br><br> 15 3/5 - 3 2/7

artcher [175]

Answer:

12 11/35

Step-by-step explanation:

Given Data

First term= 15 3/5

To simple fracrion= 78/5

Second term= 3 2/7

To simple fracrion= 23/7

Hence the operation goes thus

= 78/5- 23/7

LCM = 35

=546-115/35

=431/35

=12 11/35

6 0

3 years ago

Need help on this calculus

N76 [4]

$\displaystyle\int\sin^3t\cos^3t\,\mathrm dt$

One thing you could do is to expand either a factor of $\sin^2t$ or $\cos^2t$ , then expand the integrand. I'll do the first.

You have

$\sin^2t=1-\cos^2t$

which means the integral is equivalent to

$\displaystyle\int\sin t(1-\cos^2t)\cos^3t\,\mathrm dt$

Substitute $u=\cos t$ , so that $\mathrm du=-\sin t\,\mathrm dt$ . This makes it so that the integral above can be rewritten in terms of $u$ as

$\displaystyle-\int(1-u^2)u^3\,\mathrm du=\int(u^5-u^3)\,\mathrm du$

Now just use the power rule:

$\displaystyle\int(u^5-u^3)\,\mathrm du=\dfrac16u^6-\dfrac14u^4+C$

Back-substitute to get the antiderivative back in terms of $t$ :

$\dfrac16\cos^6t-\dfrac14\cos^4t+C$

4 0

3 years ago

Read 2 more answers

Lexie has two cans of tennis balls there are free tennis balls in each can she buys two more cans how many tennis balls does he

jasenka [17]

Okay, so if each can has 3 balls, add up each can. he has two cans right now meaning that he has 6 balls. What is 6 + 3 + 3 = ?

6 0

3 years ago

A rock is thrown upward with an initial velocity of 16 ft/s from an initial height of 5 ft. How much time does it take for the r

Bogdan [553]

Check the picture below.

so it hits the ground when y = 0, thus

$\bf ~~~~~~\textit{initial velocity}\\\\
\begin{array}{llll}
~~~~~~\textit{in feet}\\\\
h(t) = -16t^2+v_ot+h_o
\end{array} 
\quad 
\begin{cases}
v_o=\stackrel{16}{\textit{initial velocity of the object}}\\\\
h_o=\stackrel{5}{\textit{initial height of the object}}\\\\
h=\stackrel{}{\textit{height of the object at "t" seconds}}
\end{cases}$

$\bf h(t)=-16t^2+16t+5\implies \stackrel{h(t)}{0}=-16t^2+16t+5
\\\\\\
16t^2-16t-5=0\implies (4t+1)(4t-5)=0\\\\
-------------------------------\\\\
4t+1=0\implies 4t=-1\implies t=-\cfrac{1}{4}\\\\
-------------------------------\\\\
4t-5=0\implies 4t=5\implies t=\cfrac{5}{4}\implies \boxed{t=1\frac{1}{4}}$

since "t" is seconds it took, it can't be a negative amount.

5 0

3 years ago