Answer:
12 km/h
Step-by-step explanation:
Because they are asking for Pedro's average speed, you can divide the distance by the amount of time taken to travel. So in this case, Pedro's speed is equal to 96km/8h, which when divided simplifies to 12km/h. A little tip is to look at the units the answer is asking for (like in this problem it asks what was Pedro's average speed in km per hour). Km per hour is km/h, which will give you a hint on what units you will need to divide by. Hope this helps!
The circle with center O has two chords AC and EF which are of same length 9.07.
OD and OB are the two perpendiculars drawn from the center O to the two chords AC and EF .It represents the distance of the chords from the centre.
The circle theorem states: congruent chords are equidistant from the center.
OD is congruent to OB.
Option A is the right answer.
<span>Line graphs are used to display data or information that changes continuously over time. hope that helped</span>
Definition of eo-primes or relatively primes: Two numbers are said to be co-prime or relatively prime If their HCF IS 1 Hence to prove 847 and 2160 as co-prime numbers we will find their HCF and which should be 1
New steps to find HCF will be as under
2160 = 847 x 2+ 466
847 = 466 ×1 +381
466 = 381 x 1 + 85
381 =85 x 4+ 41
85 =41 x 2+3
41 =3 x 13+ 2
3 =2 x 1+1
2 =1 x 2+0
Therefore, the HCF=1 Hence, the numbers are co-primes (relatively prime).
Hmm if I'm not mistaken, is just an "ordinary" annuity, thus
![\bf \qquad \qquad \textit{Future Value of an ordinary annuity} \\\\ A=pymnt\left[ \cfrac{\left( 1+\frac{r}{n} \right)^{nt}-1}{\frac{r}{n}} \right] \\\\\\](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Cqquad%20%5Ctextit%7BFuture%20Value%20of%20an%20ordinary%20annuity%7D%0A%5C%5C%5C%5C%0AA%3Dpymnt%5Cleft%5B%20%5Ccfrac%7B%5Cleft%28%201%2B%5Cfrac%7Br%7D%7Bn%7D%20%5Cright%29%5E%7Bnt%7D-1%7D%7B%5Cfrac%7Br%7D%7Bn%7D%7D%20%5Cright%5D%0A%5C%5C%5C%5C%5C%5C)
![\bf \begin{cases} A= \begin{array}{llll} \textit{original amount}\\ \textit{already compounded} \end{array} & \begin{array}{llll} \end{array}\\ pymnt=\textit{periodic payments}\to &400\\ r=rate\to 10\%\to \frac{10}{100}\to &0.1\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{annually, so once} \end{array}\to &1\\ t=years\to &14 \end{cases} \\\\\\ A=400\left[ \cfrac{\left( 1+\frac{0.1}{1} \right)^{1\cdot 14}-1}{\frac{0.1}{1}} \right] ](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0AA%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Boriginal%20amount%7D%5C%5C%0A%5Ctextit%7Balready%20compounded%7D%0A%5Cend%7Barray%7D%20%26%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%0A%5Cend%7Barray%7D%5C%5C%0Apymnt%3D%5Ctextit%7Bperiodic%20payments%7D%5Cto%20%26400%5C%5C%0Ar%3Drate%5Cto%2010%5C%25%5Cto%20%5Cfrac%7B10%7D%7B100%7D%5Cto%20%260.1%5C%5C%0An%3D%0A%5Cbegin%7Barray%7D%7Bllll%7D%0A%5Ctextit%7Btimes%20it%20compounds%20per%20year%7D%5C%5C%0A%5Ctextit%7Bannually%2C%20so%20once%7D%0A%5Cend%7Barray%7D%5Cto%20%261%5C%5C%0A%0At%3Dyears%5Cto%20%2614%0A%5Cend%7Bcases%7D%0A%5C%5C%5C%5C%5C%5C%0AA%3D400%5Cleft%5B%20%5Ccfrac%7B%5Cleft%28%201%2B%5Cfrac%7B0.1%7D%7B1%7D%20%5Cright%29%5E%7B1%5Ccdot%2014%7D-1%7D%7B%5Cfrac%7B0.1%7D%7B1%7D%7D%20%5Cright%5D%0A)
