"Nina and Ryan each ran at a constant speed for a 100-meter race. Each runner’s distance for the same section of the race is displayed..." Ryan had a head start of 10 meters.
This is further explained below.
<h3>Who had a head start, and how big was the head start?</h3>
Generally, To help low-income children and their families, the United States government funds a program called Head Start, which offers a variety of services including early childhood education, health care, nutrition education, and parent engagement.
In conclusion, who had a head start, and how big the head start was is; Ryan had a head start of 10 meters.
Read more about distance
brainly.com/question/15172156
#SPJ1
Hello there! The answer is the first option, 31/55.
To solve this, we don't even need to do any math. Note that fractions with the same numerator and denominator will be equal to 1.
Knowing this and looking at our second and fourth options, 55/55 x 111 and 31/31 x 111, these problems are the same as 1 x 111, which results in 111, which is not less than, leaving us with the third and first options.
The third option is 55/31, meaning we have more than a whole, so we are multiplying by a number greater than 1, making our answer over 111 and this option not correct.
This leaves the first option as your answer!
If 3 inch....................................20 ft
?inc...........................................57 feet
(57*3)/20=171/20=8.55 inch
Answer:
A=115.6
Step-by-step explanation:
Hey There!
So the first step is to find the formula for area of a trapezoid
![A = \frac{1}{2} h(b1 +b2)](https://tex.z-dn.net/?f=A%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20h%28b1%20%2Bb2%29)
h = height
b1 = base 1
b2 = base 2
now all we do its plug in the values
![A=\frac{1}{2} 6.8(28.3+5.7)](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B1%7D%7B2%7D%206.8%2828.3%2B5.7%29)
![6.8x26.3=192.44\\5.7x6.8=38.76\\38.76+192.44=231.2\\\frac{231.2}{2} =115.6\\A=115.6](https://tex.z-dn.net/?f=6.8x26.3%3D192.44%5C%5C5.7x6.8%3D38.76%5C%5C38.76%2B192.44%3D231.2%5C%5C%5Cfrac%7B231.2%7D%7B2%7D%20%3D115.6%5C%5CA%3D115.6)
Hope this helps!!
Any odd multiple of 3 would be the counterexample of the given conjecture.
That is 3(2n - 1), for any natural number n, will be divisible by 3 but not divisible by 6 as 2 is not a factor.
<u>Examples:</u>
3(3) = 9
3(5) = 15
3(7) = 21 are some examples.