recalling that d = rt, distance = rate * time.
we know Hector is going at 12 mph, and he has already covered 18 miles, how long has he been biking already?
so Hector has been biking for those 18 miles for 3/2 of an hour, namely and hour and a half already.
then Wanda kicks in, rolling like a lightning at 16mph.
let's say the "meet" at the same distance "d" at "t" hours after Wanda entered, so that means that Wanda has been traveling for "t" hours, but Hector has been traveling for "t + (3/2)" because he had been biking before Wanda.
the distance both have travelled is the same "d" miles, reason why they "meet", same distance.
![\bf \begin{array}{lcccl} &\stackrel{miles}{distance}&\stackrel{mph}{rate}&\stackrel{hours}{time}\\ \cline{2-4}&\\ Hector&d&12&t+\frac{3}{2}\\[1em] Wanda&d&16&t \end{array}\qquad \implies \begin{cases} \boxed{d}=(12)\left( t+\frac{3}{2} \right)\\[1em] d=(16)(t) \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Barray%7D%7Blcccl%7D%20%26%5Cstackrel%7Bmiles%7D%7Bdistance%7D%26%5Cstackrel%7Bmph%7D%7Brate%7D%26%5Cstackrel%7Bhours%7D%7Btime%7D%5C%5C%20%5Ccline%7B2-4%7D%26%5C%5C%20Hector%26d%2612%26t%2B%5Cfrac%7B3%7D%7B2%7D%5C%5C%5B1em%5D%20Wanda%26d%2616%26t%20%5Cend%7Barray%7D%5Cqquad%20%5Cimplies%20%5Cbegin%7Bcases%7D%20%5Cboxed%7Bd%7D%3D%2812%29%5Cleft%28%20t%2B%5Cfrac%7B3%7D%7B2%7D%20%5Cright%29%5C%5C%5B1em%5D%20d%3D%2816%29%28t%29%20%5Cend%7Bcases%7D)
<span>The greatest number that has one comma when written with numerals and with number names is 999,999
In number names form is is written: nini hundred and ninety nine thousand, nine hundred and ninety nine.
</span>
Answer:
Step-by-step explanation:
The equation shown in the question can be seen graphed in the image attached below. As you can see with the graphed equation the variable x can be any real number except for -1. This is because a -1 would cause the denominator of the fraction to be equal to 0, and a fraction with a denominator as 0 is a null value and does not exist.
Answer:
Step-by-step explanation:
By using the cos square identity in trigonometry i.e., cos2ϴ = 1 – sin2 ϴ, we can evaluate the exact value of cos(33 ). For calculating the exact value of cos(∏/6), we have to substitute the value of sin(30°) in the same formula.
cos(30°) = √1 – sin230°
The value of sin30° is 1/2 (Trigonometric Ratios)
cos(30°) = √1 – (1/2)2
cos(30°) = √1 – (1/4)
cos(30°) = √(1 * 4 – 1)/4
cos(30°) = √(4 – 1)/4
cos(30°) = √3/4
Therefore, cos(30°) = √3/2
