The cycle route for the triathlon is of 26 miles
A triathlon is an athletic event which involves races which involves of swimming, cycling, and running over various distances
According to the given data,
The most common triathlon distance for Bicycle route is 40 kilometers
1 mile = 1 .6 kilometer
Now we know that 1.6 kilometer is 1 mile
So we are required to convert kilometers of triathlon race into miles
Therefore, 40 kilometers in mile will be= 40/1.6 = 25 miles
Hence the cycle route is of 26 miles .
For further reference on conversion of miles to kilometer:
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Answer:
Step-by-step explanation:
If you added all the fallowing planets mercury Venus and earth together itwould be greater than the distance from Neptune to the sun neptune is 2.759 billion miles away from the sun and the three others combined are 196.18 million.
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Answer:
The required expression is 
Step-by-step explanation:
Let the number be x
We are supposed to write the algebraic expression of given statement
Double the number = 2x
Quotient of five and two =
Now we are given that Double the sum of a double the number and quotient of five and two
So, Sum of a double the number and quotient of five and two =
So, Double the sum of a double the number and quotient of five and two =
Hence The required expression is
An equation in the form of slope intercept is written as y=mx+b. That means that we need to get y positive, which we can do by adding 3y to each side, and then subtracting 3 from each side to get y by itself. Doing all of these steps, the equation becomes 3y=5x-3. But, we're not done because we still need to get y equal to one, which we can do by dividing each side of the equation by 3. This makes the equation become y=5/3x-1.
The graph of the equation looks like this:
Answer:
a) there is s such that <u>r>s</u> and s is <u>positive</u>
b) For any <u>r>0</u> , <u>there exists s>0</u> such that s<r
Step-by-step explanation:
a) We are given a positive real number r. We need to wite that there is a positive real number that is smaller. Call that number s. Then r>s (this is equivalent to s<r, s is smaller than r) and s is positive (or s>0 if you prefer). We fill in the blanks using the bold words.
b) The last part claims that s<r, that is, s is smaller than r. We know that this must happen for all posirive real numbers r, that is, for any r>0, there is some positive s such that s<r. In other words, there exists s>0 such that s<r.
