Answer:
Top 3:
- Neon orange
- Sky blue
- Black
Step-by-step explanation:
Okie, let your friend know if that's actually true. (no offense to you I promise, it's just had people who said it was for a friend and it wasn't.)
I think the answer is 7.6m
The decimal equivalent of 45⁰ 32’ 55” is 45.55⁰
<h3>Converting degrees into decimal</h3>
The given degree is:
45⁰ 32' 55''
This can be converted to decimal as shown below
![45+\frac{32}{60} +\frac{55}{3600}](https://tex.z-dn.net/?f=45%2B%5Cfrac%7B32%7D%7B60%7D%20%2B%5Cfrac%7B55%7D%7B3600%7D)
This can be simplified further as:
![45^o32'55'' = 45+0.53+0.0153\\\\45^o32'55'' = 45.55^0](https://tex.z-dn.net/?f=45%5Eo32%2755%27%27%20%3D%2045%2B0.53%2B0.0153%5C%5C%5C%5C45%5Eo32%2755%27%27%20%3D%2045.55%5E0)
Therefore, the decimal equivalent of 45⁰ 32’ 55” is 45.55⁰
Learn more on degrees to decimal conversion here: brainly.com/question/24226195
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Answer:
1. After rearranging the formula as 's' i.e speed is the subject will be given as
.
2. The average speed of a truck that travels a journey of 748 km in 11.5 hours is 65.04 kilometre per hour.
Step-by-step explanation:
Ans 1. subject means it come first in the sentence, hence here question is like write speed first in the formula so we have
![speed = \frac{distance}{time} \\ss = \frac{dd}{tt}](https://tex.z-dn.net/?f=speed%20%3D%20%5Cfrac%7Bdistance%7D%7Btime%7D%20%5C%5Css%20%3D%20%5Cfrac%7Bdd%7D%7Btt%7D)
Ans 2.
Given :
Total distance of a journey = 748 km
Total time taken for the journey = 11.5 h
To Find:
Average speed =?
Solution :
average speed is defined as of total distance covered divided by total time taken.
So, we have
![\textrm{average speed} = \frac{\textrm{total distance covered}}{\textrm{total time taken to complete total distance}} \\= \frac{748}{11.5} \\= 65.04\ km/h](https://tex.z-dn.net/?f=%5Ctextrm%7Baverage%20speed%7D%20%3D%20%5Cfrac%7B%5Ctextrm%7Btotal%20distance%20covered%7D%7D%7B%5Ctextrm%7Btotal%20time%20taken%20to%20complete%20total%20distance%7D%7D%20%5C%5C%3D%20%5Cfrac%7B748%7D%7B11.5%7D%20%5C%5C%3D%2065.04%5C%20km%2Fh)