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cestrela7 [59]
1 year ago
6

A 16\dfrac1216 2 1 ​ 16, start fraction, 1, divided by, 2, end fraction kilometer stretch of road needs repairs. Workers can rep

air 2\dfrac142 4 1 ​ 2, start fraction, 1, divided by, 4, end fraction kilometer of road per week. How many weeks will it take to repair this stretch of road?
Mathematics
1 answer:
Solnce55 [7]1 year ago
7 0

Answer:

It will take  7\frac{1}{3} weeks to complete the repair of the stretch of road.

Step-by-step explanation:

Given:

Length of the roads that need repairs = 16\frac{1}{2}\ km

Length of the road workers can repair in a week = 2\frac{1}{4}\ km

To find the number of weeks will it take to repair this stretch of road.

Solution:

Unit rate of workers to repair  roads =   2\frac{1}{4}\ km per week.

Total length of the roads to repair = 16\frac{1}{2}\ km

The number of weeks it will take to repair  16\frac{1}{2}\ km of road by workers can be given as:

⇒ \frac{16\frac{1}{2}\ km}{2\frac{1}{4}\ km}

<em>In order to divide mixed numbers, we convert them first to fractions.</em>

⇒ \frac{\frac{33}{2}} {\frac{9}{4}}

<em>To divide fractions, we take reciprocal of the divisor and replace division with multiplication:</em>

⇒  \frac{33}{2}\times \frac{4}{9}

⇒ \frac{22}{3}

We reconvert the fraction to mixed number.

⇒ 7\frac{1}{3} weeks

Thus, it will take  7\frac{1}{3} weeks to complete the repair of the stretch of road.

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Step-by-step explanation:

Let assume the dealer sold the bottle now for $P, then invested that money at 5% interest. The return would be:

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P will cancel out

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7 0
2 years ago
What is the domain and range of the equation y=500(1.08)^x
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6 0
2 years ago
Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val
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Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

A(t) = Pe^{rt}

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This means that P = 330

$590 in the year 2007

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We use this to find r.

A(t) = Pe^{rt}

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Applying ln to both sides:

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r = \frac{\ln{1.79}}{8}

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Determine the value of the account, to the nearest dollar, in the year 2009.

2009 is 10 years after 1999, so this is A(10).

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A(10) = 330e^{0.0726*10} = 682

The value of the account in the year 2009 will be $682.

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2 years ago
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<span>The perpendicular bisector of a segment is a line that cuts the segment into two equal parts (bisector) and that forms with the segment a right angle (perpendicular). Any point on the perpendicular bisector has the same distance from the segment's extremities. PC has exactly the characteristics of a perpendicular bisector of AB. </span>
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