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labwork [276]
3 years ago
5

Phone Company A charges a monthly fee of $42.50, and $0.02 for each minute talk time. Phone company B charges a monthly fee of $

25.00, and $0.09 for each minute of talk time.
Mathematics
1 answer:
Y_Kistochka [10]3 years ago
8 0
We need to call for x minute
+ Phone Company A charges a monthly fee of $42.50, and $0.02 for each minute talk time. So we have to spend: <span>$42.50+ $0.02x
+ </span>Phone company B charges a monthly fee of $25.00, and $0.09 for each minute of talk time. So we have to spend: <span>$25.00+ $0.09x

We solve for x: </span>$42.50+ $0.02x> <span>$25.00+ $0.09x
or </span>$42.50- $25.00 > $0.09x- <span>$0.02x
and we have $0.07x<$27.50
or x< 27.50:0.07 and x< 393.86

The answer is:
If we have to call much time, at least 394 minutes, we should choose A
If not, choose B</span>
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\dfrac{\mathrm d^2y}{\mathrm dx^2}=-\dfrac{3+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}y=-\dfrac{3+\frac{9x^2}{y^2}}y=-\dfrac{3y^2+9x^2}{y^3}

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