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3 years ago

Find the surface area of the cube (5.1cm tall)

Mathematics

2 answers:

lawyer [7]3 years ago

8 0

156.06 centimeters is the answer

joja [24]3 years ago

6 0

Answer:

156.06cm

Step-by-step explanation:

sa=LW6

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Find the product. (-2x^2)^3 ·3x

poizon [28]

(-2x^2)^3 * 3x

3(-8(x^2)^3)x

3(-8x^6)x

3(-8x^7)

-24x^7

Hope this is accurate & helpful! Let me know if I made any errors or if you have any additional questions.

3 0

2 years ago

Lesson 16

Goshia [24]

Answer:

546 ft³

Step-by-step explanation:

In the figure attached, the top view of the pool is shown. It can be decomposed into a trapezoid which height is 12 ft and which bases are 9 ft and 11 ft; and a rectangle which height is 1.5 ft and and it base is 11 ft.

Area of the trapezoid: (9 + 11)/2 * 12 = 120 ft²

Area of the rectangle: 1.5*11 = 16.5 ft²

Total area: 120 + 16.5 = 136.5 ft²

Volume of the pool: Total area * deep = 136.5 * 4 = 546 ft³

6 0

3 years ago

Together, two wildlife refuge tour guides lead nine 45-minute tours each day. The wildlife refuge needs to calculate the person-

andrew-mc [135]

C 13.5 bubble id the best choice

7 0

3 years ago

The pair of square pyramids are similar. Use the given information to find the scale factor of the smaller square pyramid to the

Mrrafil [7]

$\bf \qquad \qquad \textit{ratio relations}
\\\\
\begin{array}{ccccllll}
&\stackrel{ratio~of~the}{Sides}&\stackrel{ratio~of~the}{Areas}&\stackrel{ratio~of~the}{Volumes}\\
&-----&-----&-----\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}
\end{array}\\\\
-----------------------------$

$\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\
-------------------------------\\\\
\stackrel{\stackrel{pyramids}{scale~factor}}{\cfrac{small}{large}}\qquad \qquad \cfrac{s}{s}=\cfrac{\sqrt[3]{64}}{\sqrt[3]{343}}\implies \cfrac{s}{s}=\cfrac{4}{7}\implies 4:7$

8 0

3 years ago

Y = 3 sin2x, y = 0, 0 ≤ x ≤ π; about the x−axis

Dominik [7]

I assume you're revolving the region with those bounds about the x-axis, and supposed to find the volume.

Via the disk method,

$\displaystyle\pi\int_0^\pi(3\sin2x)^2\,\mathrm dx=9\pi\int_0^\pi\sin^22x\,\mathrm dx$

Recall the half-angle identity for sine:

$\sin^2t=\dfrac{1-\cos2t}2$
$\implies\displaystyle\frac{9\pi}2\int_0^\pi(1-\cos4x)\,\mathrm dx$
$=\displaystyle\frac{9\pi}2\left(x-\frac14\sin4x\right)\bigg|_{x=0}^{x=\pi}$
$=\dfrac{9\pi^2}2$

8 0

3 years ago