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2 years ago

Find the volume of the pyramid. Round to the nearest tenth if necessary. PLEASE HELP!!

Mathematics

2 answers:

harina [27]2 years ago

6 0

Answer:

1/2 x 4.1 x 2.6 = 5.33

5.33 x 5.1 = 27.183

Rounding off to nearest tenth = 27.2 m^3

jok3333 [9.3K]2 years ago

5 0

Answer:

27.183, 27.2

Step-by-step explanation:

First, do the formula for the triangle:

$\frac{1}{2}$ x 4.1 x 2.6 = 5.33

Next, multiply that by the base side:

5.33 x 5.1 = 27.183

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HELP!!!<br> 50 Points!!!!!

andreyandreev [35.5K]

Answer:

i think the first and last one.

Step-by-step explanation:

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2 years ago

Suppose that an angle is bisected to form two smaller angles that measure (2+30)° and (8+6)° . What is the measure of the origin

Valentin [98]

M<A=(2+30) M<B=(8+6)

(2+30)+(8+6)

32+14

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Answer: The original angle measured 46 Degrees

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3 years ago

The radius of a spherical balloon is measured as 20 inches, with a possible error of 0.03 inch. Use differentials to approximate

soldi70 [24.7K]

Answer:

a) $V = 33510.322\,in^{3}$ , b) $A_{s} = 5026.548\,in^{2}$ , c) $\% V = 0.450\,\%$ , $\%A_{s} = 0.300\,\%$ .

Step-by-step explanation:

The volume and the surface area of the sphere are, respectively:

$V = \frac{4}{3}\pi \cdot r^{3}$

$A_{s} = 4\pi \cdot r^{2}$

a) The volume of the sphere is:

$V = \frac{4}{3}\pi \cdot (20\,in)^{3}$

$V = 33510.322\,in^{3}$

b) The surface area of the sphere is:

$A_{s} = 4\pi \cdot (20\,in)^{2}$

$A_{s} = 5026.548\,in^{2}$

c) The total differentials for volume and surface area of the sphere are, respectively:

$\Delta V = 4\pi\cdot r^{2}\,\Delta r$

$\Delta V = 4\pi \cdot (20\,in)^{2}\cdot (0.03\,in)$

$\Delta V = 150.796\,in^{3}$

$\Delta A_{s} = 8\pi\cdot r \,\Delta r$

$\Delta A_{s} = 8\pi \cdot (20\,in)\cdot (0.03\,in)$

$\Delta A_{s} = 15.080\,in^{2}$

Relative errors are presented hereafter:

$\%V = \frac{\Delta V}{V}\times 100\%$

$\%V = \frac{150.796 \,in^{3}}{33510.322\,in^{3}}\times 100\,\%$

$\% V = 0.450\,\%$

$\% A_{s} = \frac{\Delta A_{s}}{A_{s}}\times 100\,\%$

$\% A_{s} = \frac{15.080\,in^{2}}{5026.548\,in^{2}}\times 100\,\%$

$\%A_{s} = 0.300\,\%$

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3 years ago

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