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3 years ago

Given uv = su tu sv prove vt = st

Mathematics

1 answer:

Lady bird [3.3K]3 years ago

5 0

Answer: see proof below

<u>Step-by-step explanation:</u>

$\begin{array}{cl|cl}&\underline{\text{Statement}}&&\underline{\text{Reason}}\\1.&TU\perp SV&1.&\text{Given}\\2.&\angle TUV = 90^o&2.&\text{De finition of Perpendicular angles}\\&\angle TUS = 90^o&&\\3.&\angle TUV=\angle TUS&3.&\text{Transitive Property}\\4.&UV=US&4.&\text{Given}\\5.&TU=TU &5.&\text{Reflexive Property}\\6.&\triangle TUV = \triangle TUS&6.&\text{SAS Congruency Theorem}\\7.&VT=ST&7.&CPCTC\end{array}$

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Approximate the change in the volume of a sphere when its radius changes from r = 40 ft to r equals 40.05 ft (Upper V (r )equal

alexgriva [62]

Answer:

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

Step-by-step explanation:

The volume of the sphere ( $V$ ), measured in cubic feet, is represented by the following formula:

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

Where $r$ is the radius of the sphere, measured in feet.

The change in volume is obtained by means of definition of total difference:

$\Delta V = \frac{\partial V}{\partial r}\Delta r$

The derivative of the volume as a function of radius is:

$\frac{\partial V}{\partial r} = 4\pi \cdot r^{2}$

Then, the change in volume is expanded:

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

If $r = 40\,ft$ and $\Delta r = 40\,ft-40.05\,ft = 0.05\,ft$ , the change in the volume of the sphere is approximately:

$\Delta V \approx 4\pi\cdot (40\,ft)^{2}\cdot (0.05\,ft)$

$\Delta V \approx 1005.310\,ft^{3}$

The change in the volume of a sphere whose radius changes from 40 feet to 40.05 feet is approximately 1005.310 cubic feet.

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3 years ago

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2 years ago

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Answer:

Step-by-step explanation:

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