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omeli [17]
3 years ago
10

Madu. For every N3.00 Abam put in,

Mathematics
1 answer:
pav-90 [236]3 years ago
8 0

Answer:

A:E:M = 6:4:5

Step-by-step explanation:

Given

Abam =N3.00 when Emeka = N2.00

Emeka = N4.00 when Madu = N5.00

Required

Determine their ratios

Represent their ratios with the first letter of their names:

From the first statement

A:E = 3:2 ---- (1)

From the second

E:M = 4:5 ----- (2)

Multiply the ratio in (1) by 2

A:E = 3 * 2: 2 * 2

A:E = 6: 4 ---- (3)

Notice that the value of E in both cases is the same i.e E = 4

(1) and (3) can be merged to give:

A:E:M = 6:4:5

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Read 2 more answers
1. Differentiate with respect to x:<br>​
natta225 [31]

9514 1404 393

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  a) y' = x^2(3x·ln(6x) +1)

  b) y' = 6e^(3x)/(1 -e^(3x))^2

Step-by-step explanation:

The applicable rules for derivatives include ...

  d(u^n)/dx = n·u^(n-1)·du/dx

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  d(e^u)/dx = e^u·du/dx

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__

(a)

  y=x^3\ln{(6x)}\\\\y'=3x^2\ln{(6x)}+\dfrac{x^3\cdot6}{6x}\\\\\boxed{\dfrac{dy}{dx}=3x^3\ln{(6x)}+x^2}

__

(b)

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3 years ago
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