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ValentinkaMS [17]
4 years ago
13

The quote for USD/CHF is listed as 1.4481/88. How many swiss francs does it cost to buy one dollar?

Mathematics
1 answer:
pychu [463]4 years ago
4 0
Based on this quote, this answer is 60.77.

The ratio given in USD to CHF is 1.4481/88.  Since we want 1 USD, we divide both parts of the ratio by 1.4481:

1.4481 ÷ 1.4481 = 1

88 ÷ 1.4481 = 60.77
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A) For finding the height of the rocket 3 seconds after the launch, we will <u>plug t=3 into the above equation</u>. So....

H= -16(3)^2+500(3)+20\\ \\ H= -16(9)+1500+20\\ \\ H= -144+1500+20=1376

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B) When the rocket at a height of 400 feet, then <u>we will plug H= 400</u>

400=-16t^2+500t+20\\ \\ 16t^2-500t-20+400=0\\ \\ 16t^2-500t+380=0\\ \\ 4(4t^2-125t+95)=0\\ \\ 4t^2-125t+95=0

Using quadratic formula, we will get......

t= \frac{-(-125)+/-\sqrt{(-125)^2-4(4)(95)}}{2(4)}\\ \\ t= \frac{125+/-\sqrt{14105}}{8}\\ \\ t= 30.4705... \\ and \\ t= 0.7794...

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C) The time duration that the rocket remains in the air means we need to find <u>the time taken by the rocket to reach the ground</u>. When it reaches the ground, then H=0. So.....

0=-16t^2+500t+20\\ \\ -4(4t^2-125t-5)=0\\ \\ 4t^2-125t-5=0

Using <u>quadratic formula</u>, we will get.....

t= \frac{-(-125)+/-\sqrt{(-125)^2-4(4)(-5)}}{2(4)}\\ \\ t= \frac{125+/-\sqrt{15705}}{8}\\ \\ t=31.2899...\\ and\\ t= -0.0399...

<em>(Negative value is ignored as time can't be in negative)</em>

So, the rocket will remain in the air for 31.2899... seconds.

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Step-by-step explanation:

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follow these steps:

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Cut out the squares. Place each square next to the corresponding sides of the triangle.

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The total area of a big square, where each side having a length of a+b, is:

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