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Anna35 [415]
3 years ago
8

on a suspension bridge the roadway is hung from cables hanging between support towers. the cable of one bridge segment of the pa

rabola y=.1x^2-7x +150 where y is the height of the cable in feet above the roadway at the distance x feet from a support tower. what is the closest the cable comes to the roadway? how far from the support towers does this occur
Mathematics
1 answer:
Greeley [361]3 years ago
5 0
<span>The parabola opens upward and is symmetric to the y-axis. Its general form is: . y \;=\;ax^2 + c Its y-intercept is (0, 10) . . . Hence: . y \;=\;ax^2 + 10 It passes through (200, 100). We have: . 100 \:=\:a\cdot200^2 + 10 \quad\Rightarrow\quad a \:=\:\frac{9}{4000} Hence: . y \;=\;\tfrac{9}{4000}x^2 + 10 When x = \pm50,\;\;y \:=\:\tfrac{9}{4000}(50^2) + 10 \:=\:\frac{125}{8} Therefore, 50 feet from the center, the cable is 15\tfrac{5}{8} feet high.</span>
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D. $31,337.27

Step-by-step explanation:

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Also, it is given that the loan was compounded annually.

We have the formula as,

P=\frac{\frac{r}{n}\times PV}{1-(1+\frac{r}{n})^{-t\times n}}

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Substituting the values, we get,

i.e. PV=\frac{P\times [1-(1+\frac{0.075}{12})^{-10\times 12}]}{\frac{0.075}{12}}

i.e. 22000=\frac{P\times [1-(1+0.00625)^{-120}]}{0.00625}

i.e. 22000=\frac{P\times [1-(1.00625)^{-120}]}{0.00625}

i.e. 22000=\frac{P\times [1-0.4735]}{0.00625}

i.e. 22000=\frac{P\times 0.5265}{0.00625}

i.e. P=\frac{22000\times 0.00625}{0.5265}

i.e. P=\frac{137.5}{0.5265}

i.e. P=261.16

Thus, the total lifetime cost to pay of the loans compounded annually  = 261.16 × 120 = $31,339.2

Hence, the total cost close to the answer is $31,337.27

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