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muminat
3 years ago
14

An athlete training for the Olympics women's each lap for a 4 lap race in the following times: 54.73, 54.56, and 54.54 seconds.

What is the total time it takes her to swim the four laps.
Mathematics
1 answer:
natita [175]3 years ago
4 0

Answer:

The total time it takes her to swim the four laps is 3.65 minutes.

Step-by-step explanation:

Here, according to the given data:

The time taken to complete 1st lap =  54.73 seconds

The time taken to complete 2nd lap =  54.56 seconds

The time taken to complete 3rd lap =  54.32 seconds

The time taken to complete 4th lap =  54.54 seconds

Now, the TOTAL TIME taken to swim all 4 laps  = SUM OF ALL 4 TIMES

= (54.73  + 54.56 + 54.32 + 54.54) seconds

= 219.35 Seconds

Now, 60 seconds = 1 minute

⇒ 1 seconds  = 1/60 minute

⇒219.35 seconds = 219.27 x (1/60) minutes = 3.6558  minutes

Hence, the total time it takes her to swim the four laps is 3.65 minutes.

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Which of the following statements accurately describes the expression x+5/x^2-2
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Answer:

C

Step-by-step explanation:

It’s how you would solve the equation

5 0
3 years ago
3) Smurfette is glad you're getting on with your adventure, but she wants to warn you that this last
madreJ [45]

Answer:

take x=2.515834 and r=14.31386

Step-by-step explanation:

Let us call the the lenght of the side of the square as x. The total lenght of the wire is 100. Note that if we have a square of side x, it has a perimeter of 4x. That means, that we need a wire of length 4x to build a square of side x. Note that if we take a wire of length 4x, we have that the remainder is 100-4x. We will use this to form a circle. As for the square, since we are using the wire of length 100-4x to build a circle, it must happen that the perimeter of the circle is 100-4x.

Recall that the perimeter of a circle of radius r is 2\pi r, so in this case , we have that 2\pi r = 100-4x which implies that r =\frac{100-4x}{2\pi}.

Now, recall that the area of a square of side x is x^2. Also, recall that the area of a circle of radius r is \pi r^2. In our case, the area of the circle is given by

\pi r^2 = \pi (\frac{100-4x}{2\pi})^2

Also, we are given that the sum of the areas of both figures is 650. That is

x^2 + \pi (\frac{100-4x}{2\pi})^2 = x^2+\frac{(100-4x)^2}{4\pi}=650

By substracting 650 on both sides and multiplying by 4\pi this equation is equivalent to

4\pi x^2 + (100-4x)^2-650\cdot 4\pi =0

if we expand (100-4x)^2 and then group each term of x, x^2 we get :

(4\pi+16)x^2-800x+100^2-650\cdot 4\pi =0

Recall that given a cuadractic equation of the form ax^2+bx+c=0 the solution is given by

x = \frac{-b \pm \sqrt[]{b^2-4ac}}{2a}

In this case, the existence of a real solution is given by the expression b^2-4ac. For it to have a real solution, it must happen that b^2-4ac\geq 0

In this case, b=-800, c = 100^2-650\cdot 4\pi, a = 4\pi+16[/tex]

For this values, we get that b^2-4ac = 430681\geq 0, so the is a solution for our problem.

In this case,

x = \frac{-b - \sqrt[]{b^2-4ac}}{2a}= 2.515834

and x = \frac{-b + \sqrt[]{b^2-4ac}}{2a}= 25.4891 are the two possible solutions for this problem. We can check that this values of x fulfill our restrictions. Note that if x=25.4891, by replacing in the value of r, we get that r=-0.3113853. Since r is the radius of the circle, it must be positive. Hence, x=25.4891 is discarted.For x=2.515834 we get a value of r=14.31386, so this is the answer

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S_A_V [24]

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