I'm assuming 43 is actually 4/3
see what number the previous term must be multiplied by to get the next term
4 *something = 12, 12*something=36
common ratio is 3
4x = (130 - 2x) so x = 65/3
(3y + 40) + (130 -2x) = 180 because the angles are supplementary
plug in x
3y + 40 + 130 - 2(65/3) = 180
so 3y = 95/3
therefore y = 95/6
Im guessing either 2/3*570 or 1/3*5700
The given operation involves just swapping the two rows, so carrying out
on the matrix gives
![\left[\begin{array}{cc|c}5&5&5\\1&-\dfrac23&5\end{array}\right]\to\left[\begin{array}{cc|c}1&-\dfrac23&5\\5&5&5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D5%265%265%5C%5C1%26-%5Cdfrac23%265%5Cend%7Barray%7D%5Cright%5D%5Cto%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Cc%7D1%26-%5Cdfrac23%265%5C%5C5%265%265%5Cend%7Barray%7D%5Cright%5D)
Answer:
He can take 5 Horses at max in his trailer at one time without going over the max weight his truck can tow. (Assuming the average weight of one horse to be equal to 1000 pounds)
Step-by-step explanation:
The no. of horses that can be carried by the truck can be found by simply dividing the maximum weight, that the truck can tow by the weight of a horse.
Max. No of Horses = (Max weight truck can tow)/(Average weight of one horse)
The weight of a horse is not given in the question .Thus, we assume the average weight of one horse, to be equal to 1000 pounds, we get:
Max. No of Horses = 5000 pounds/ 1000 pounds
<u>Max. No of Horses = 5</u>
