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aleksandrvk [35]
2 years ago
8

Find the ranges giving brainlest

Mathematics
1 answer:
scoray [572]2 years ago
5 0

Answer:

The range for Problem 18 is

16.1

The range for problem 19 is

167

Step-by-step explanation:

To find the range you subtract the smallest value from the largest value. In problem 18 the largest value was 17.6 and the smallest was 1.5.

17.6 - 1.5 = 16.1

In problem 19 the largest value was 181 and the smallest was 14

181 - 14 = 167

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A bicyclist is finishing his repair of a flat tire when a friend rides by with a constant speed of 3.63 m/s. Four seconds later,
castortr0y [4]

From the moment the friend passes the bicyclist, his friend covers a distance over time t of (3.63 m/s)*t.

The bicyclist covers a distance of 1/2*(2.11 m/s^2)*t^2. They meet when these distances are equal:

3.63 t = 1.055 t^2  ==>  1.055 t^2 - 3.63 t = 0

==>  t = 0 s   or   t = 3.44 s

5 0
3 years ago
Help me answer this.. plz
Fudgin [204]

Answer:

2\frac{1}{30}

Step-by-step explanation:

First thing you gotta do is find the common denominator.

Do this by listing factors of the denominators:

5- 5, 10, 15, 20, 25, 30

6- 6, 12, 18, 24, 30, 36

5 and 6 both have 30 in common so that's the new denominator for all 3 of the terms.

What you do to the bottom you must do to the top.

Since you multiply 5 by 6 to get 30, you must multiply 2 and 4 by 6 as well.

Since you multiply 6 by 5 to get 30, then multiply 5 by 5 as well.

The new equation is:

\frac{12}{30} +\frac{25}{30} + \frac{24}{30} =

Now add the numerators together

The solution is \frac{61}{30}. Now simplify. 61 is prime, so you cannot change the fraction How many times does 30 go into 61? Two times. This gives us 2\frac{1}{30}.

4 0
3 years ago
Do u agree?I m not pretty sure
nika2105 [10]
It looks correct to me :)
6 0
3 years ago
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