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alexandr402 [8]
3 years ago
9

What are the discontinuities of the function f(x) =x<2-36/4x-24 ?

Mathematics
1 answer:
kipiarov [429]3 years ago
4 0
Well, 2-36/4x-24 is the same as (2-24)-(36/4)x which simplifies to -22-9x. This does not have a discontinuity anywhere. If you meant to divide 36 by 4x, then x=0 would be the discontinuity since you cannot divide by 0. If you meant to have 4x-24 all in the denominator, then x=6 is the discontinuity since that would make the bottom equal 0. So it seems you need some parenthesis here somewhere for your problem to make sense.
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Since line DB is perpendicular to line AC, then arc AB = arc BC

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Collecting like terms,

14x - 7x = 18 + 3

7x = 21

Dividing both sides by 7, we have

\begin{gathered} x=\frac{21}{7} \\ x=3 \end{gathered}\begin{gathered} \angle AC=\angle AOB+\angle BOC \\ \angle AC=14x-3+7x+18 \\  \end{gathered}

substituting x into the above equation,

\begin{gathered} \angle AC=14(3)-3+7(3)+18 \\ \angle AC=42-3+21+18 \\ \angle AC=78^0 \end{gathered}

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1 year ago
13abc + 2a2d - 14b3c4
Setler [38]

Answer:

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3 years ago
Two similar cones have volume of 4 m and 108 m respectively. If the large one has surface area 54 m, find the surface area of th
trasher [3.6K]

 

\displaystyle\bf\\Explanations:\\\\The~similarity~ratio~of~two~similar~cones=k\\\\k=is~the~ratio~between~2~corresponding~lengths\\\\k=\frac{R_1}{R_2}=\frac{h_1}{h_2}\\\\The~ratio~between~2~corresponding~areas~of~similar~cones=k^2\\\\\frac{Area~1}{Area~2}=k^2\\\\The~ratio~between~the~volumes~of~the~2~similar~cones=k^3\\\\\frac{Volume~1}{Volume~2}=k^3

.

\displaystyle\bf\\Solving:\\\\Volume1=108~m^3\\\\Volume2=4m^3\\\\k^3=\frac{108}{4}\\\\k^3=27~~\Big|\sqrt{~}\\\\k=\sqrt[\b3]{27}\\\\k=3\\\\Area1=54~m^2\\\\\frac{Area~1}{Area~2}=k^2\\\\\frac{54}{Area~2}=3^2\\\\\frac{54}{Area~2}=9\\\\Area2=\frac{54}{9}\\\\\boxed{\bf Area2=6 m^2}

 

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erastovalidia [21]
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vitfil [10]

Answer:

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Step-by-step explanation:

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