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kramer
3 years ago
6

When it hatches from its egg, the shell of certain crab is 1cm across. When fully grown the shell is approximately 10cm across.

Each new shell is one-third bigger than the previous one. How many shell does a fully grown carb have during its life?
Mathematics
1 answer:
inn [45]3 years ago
8 0
Well every time it gets rid of a shell it grows 1 1/3 times larger. So so to see how much it grows after the first time you say
1 cm * 1 1/3 = 1 1/3 cm

to get the next one you do the same
1 1/3 cm * 1 1/3 cm = 16/9

It will keep going multiplying by 1 1/3. So so we can say in an equation that the
(initial size) * 1 1/3 *(number of shells) = length

Or 1cm * 1 1/3 * n = L

We we know the final length is 10cm

So 1 1/3 * n = 10cm

n = 7.5shells

so approximately 7 or 8 shells
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Attached Image

Step-by-step explanation:

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Y-Intercepts: (0, 2.813)

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2 years ago
I need to know if it’s A B C OR D
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SOLUTION

Consider the property, opposite angle of a cyclic quadrilateral are supplementary

hence

\angle NMP+\angle NOP=180^0

Recall that

\begin{gathered} \angle NMP=\angle M=(8x-24)^0 \\ \text{and } \\ \angle NOP=\angle O=(4x)^0 \end{gathered}

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\begin{gathered} \angle M+\angle O=180^0(opposite\text{ angles of a cyclic quadrilateral)} \\ (8x-24)^0+4x^0=180^0 \end{gathered}

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\begin{gathered} 8x-24+4x=180 \\ 8x+4x-24=180 \\ 12x-24=180 \\ \text{Add 24 t o both sides } \\ 12x-24+24=180+24 \\ 12x=204 \\ \text{divide both sides by 12} \\ \frac{12x}{12}=\frac{204}{12} \\ \text{Then} \\ x=17 \end{gathered}

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x=17

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\begin{gathered} \angle NOP=(4x)^0 \\ \text{substitute the value of x, we have } \\ \angle NOP=4\times17=68^0 \\ \text{Hence } \\ \angle NOP=68^0 \end{gathered}

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The measure of angle NOP is 68⁰

Answer; 68⁰ (The fourth Option )

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1 year ago
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Pachacha [2.7K]
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