Answer:
26.27
Step-by-step explanation:
divide its easy you can never go wrong hope you find this helpful
Midline theorem states that the cuts along the midline of a triangle creates a segment that is parallel to the base and half of the base.. The length of the line segment MN is 58 cm. Thus the option A is the correct option.
Given information-
The figure for the given problem is attached below.
In the given figure,
![AM=MB](https://tex.z-dn.net/?f=AM%3DMB)
Thus the<em> M</em> is the midpoint of the line segment <em>AB.</em>
In the given figure,
![BN=NC](https://tex.z-dn.net/?f=BN%3DNC)
Thus the<em> N</em> is the midpoint of the line segment <em>BC.</em>
<h3>Midline theorem</h3>
Midline theorem states that the cuts along the midline of a triangle creates a segment that is parallel to the base and half of the base.
Thus the length of the line segment<em> MN</em> can be given as,
![MN=\dfrac{1}{2} \times AC\\ ](https://tex.z-dn.net/?f=MN%3D%5Cdfrac%7B1%7D%7B2%7D%20%5Ctimes%20AC%5C%5C%0A)
Put the values ,
![MN=\dfrac{1}{2} \times 116\\](https://tex.z-dn.net/?f=MN%3D%5Cdfrac%7B1%7D%7B2%7D%20%5Ctimes%20116%5C%5C)
![MN=58](https://tex.z-dn.net/?f=MN%3D58)
Thus the length of the line segment MN is 58 cm. Thus the option A is the correct option.
Learn more about the midline theorem here;
brainly.com/question/7717322
Answer:
The Ballwin Bears are taller on average, and the Aviston Aces have players whose heights are more consistent.
Step-by-step explanation:
for derby dragon;
mean height = 72 inches
standard deviation = 1.2 inches
for aviston aces:
mean height = 70.8
standard deviation = 0.7 inches
for balwin bears;
mean height = 73 inches
standard deviation = 1.0 inches
the mean height of aviston aces < mean height of derby < mean height of balwin bears
So on average the balwin bears are taller.
The standard deviation of derby dragon is > that of balwin bears > aviston aces.
So the more consistent height is that of aviston aces players.
Answer:
<em>Pool 1 leaks faster than pool 2.</em>
Step-by-step explanation:
<u>Rates of change</u>
The rate of change (ROC) is a measure that compares two quantities, usually to know how fast one variable changes in time.
We are given two rates of change for two pools that are leaking. The first one loses 2/3 gallon in 15 minutes, and the other loses 3/4 gallon in 20 minutes.
To compare them, we are required to express time in hours. Recall one hour has 60 minutes, or equivalently, one minute has 1/60 hours. Converting both times, we have:
15 minutes = 15/60 = 1/4 hours
20 minutes = 20/60 = 1/3 hours
Now compute both rates of change:
Pool 1:
![\displaystyle ROC_1=\frac{2/3}{1/4}=\frac{8}{3}\approx 2.67\ gal/h](https://tex.z-dn.net/?f=%5Cdisplaystyle%20ROC_1%3D%5Cfrac%7B2%2F3%7D%7B1%2F4%7D%3D%5Cfrac%7B8%7D%7B3%7D%5Capprox%202.67%5C%20gal%2Fh)
Pool 2:
![\displaystyle ROC_2=\frac{3/4}{1/3}=\frac{9}{4}= 2.25\ gal/h](https://tex.z-dn.net/?f=%5Cdisplaystyle%20ROC_2%3D%5Cfrac%7B3%2F4%7D%7B1%2F3%7D%3D%5Cfrac%7B9%7D%7B4%7D%3D%202.25%5C%20gal%2Fh)
Comparing both ratios, it's clear pool 1 leaks faster than pool 2.
Answer: -6
-6•-7=42
Negative times a negative equals a positive