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Jovelyn580go
Answered



[tex]Radical\\Equations\\ Applied[/tex]

You work for a company that
manufactures plumb bobs. The same mould
is used to cast plumb bobs of different sizes,
The equation

[tex]h = \frac{3 \sqrt[3]{t} }{2}, 0 \leqslant h \leqslant 3[/tex]

models the relationship between the height "
(in inches) of the plumb bob and the time (in seconds) that metal
alloy is poured into the mould.
You are tasked to present to the company the length of time
to pour alloy to the mould to cast a plumb bob of different heights.
Your outputs will be the new equation which may be used to solve
for + and a table showing the required time for a specific plumb bob
height and the computations.​

TexRadicalEquations AppliedtexYou Work For A Company Thatmanufactures Plumb Bobs The Same Mouldis Used To Cast Plumb Bobs Of Different SizesThe Equationtexh Fra class=

Sagot :

AGboiigo

SOLUTION

[tex]\\[/tex]

[tex] \large \tt{given \: \: that} \: \: h = \large \bold{\orange{\frac{3 \sqrt[3]{t} }{2}}}[/tex][tex]\\[/tex]

[tex]\\[/tex]

Height = 2

[tex]2 \:= \frac{3 \sqrt[3]{t} }{2} \\ 2(2) = (\frac{3 \sqrt[3]{t} }{2})2 \\ 4 = {3 \sqrt[3]{t}} \\ (4)^{3} = {(3 \sqrt[3]{t})^{3} } \\ 64 = 27t \\ \boxed{\green{\frac{64}{27}} = t}[/tex]

[tex]\\[/tex]

Height = 2.5

[tex]2.5\:= \frac{3 \sqrt[3]{t} }{2} \\ 2(2.5) = (\frac{3 \sqrt[3]{t} }{2})2 \\ 5 = {3 \sqrt[3]{t}} \\ (5)^{3} = {(3 \sqrt[3]{t})^{3} } \\ 125 = 27t \\ \boxed{\green{\frac{125}{27}} = t}[/tex]

[tex]\\[/tex]

Height = 3

[tex]3\:= \frac{3 \sqrt[3]{t} }{2} \\ 2(3) = (\frac{3 \sqrt[3]{t} }{2})2 \\ 6 = {3 \sqrt[3]{t}} \\ (6)^{3} = {(3 \sqrt[3]{t})^{3} } \\ 216 = 27t \\ \boxed{\green{\frac{216}{27}} = t}[/tex]

[tex]\\[/tex]

ANSWER

[tex]\begin{array}{|c|c|}\hline \tt \bold{height}& \tt \bold{time} \\ \hline \tt {{2} } & \tt {2.37} \\ \hline\ {{2.5}}& \tt 4.63\\ \hline \tt {3} & \tt8 \\ \hline \end{array}[/tex]

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