Can I please have some help?

Mathematics

1 answer:

Dvinal [7]3 years ago

7 0

Hey the answer is 4000 ft^3 , choice c

You might be interested in

Need an explanation

aleksandrvk [35]

Answer: Choice B

$(6\sqrt{5}+5)i$

==========================================================

Work Shown:

$\sqrt{-5}+\sqrt{-25}+\sqrt{-125}\\\\\sqrt{-1*5}+\sqrt{-1*25}+\sqrt{-1*25*5}\\\\\sqrt{-1}*\sqrt{5}+\sqrt{-1}*\sqrt{25}+\sqrt{-1}*\sqrt{25}*\sqrt{5}\\\\i*\sqrt{5}+i*5+i*5*\sqrt{5}\\\\i*\sqrt{5}+5i+5i*\sqrt{5}\\\\(i\sqrt{5}+5i*\sqrt{5})+5i\\\\(\sqrt{5}+5\sqrt{5})i+5i\\\\(6\sqrt{5})i+5i\\\\(6\sqrt{5}+5)i\\\\$

This points us to answer choice B

7 0

3 years ago

if the perimeter of a rectangular picture frame must be less than 180 in., and the width is 26 in., what must the hight (h) of t

juin [17]

First since the perimeter of this problem is P=2b x 2h I first multiplied 26 by 2 and got 52 inches. then I subtracted 52 from 180 and got 128. From there I divide 128 by 2 and I got 64 for the length

6 0

3 years ago

Find the distance traveled by a particle with position (x, y) as t varies in the given time interval. x = 5 sin2 t, y = 5 cos2 t

rodikova [14]

$\begin{cases}x(t)=5\sin2t\\y(t)=5\cos2t\end{cases}\implies\begin{cases}\frac{\mathrm dx}{\mathrm dt}=10\cos2t\\\frac{\mathrm dy}{\mathrm dt}=-10\sin2t\end{cases}$

The distance traveled by the particle is given by the definite integral

$\displaystyle\int_C\mathrm dS=\int_0^{3\pi}\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt$

where $C$ is the path of the particle. The distance is then

$\displaystyle\int_0^{3\pi}\sqrt{100\cos^22t+100\sin^22t}\,\mathrm dt=10\int_0^{3\pi}\mathrm dt=30\pi$