<span>#include <iostream>
using namespace std;
class InventoryTag {
public:
InventoryTag();
int getQuantityRemaining() const;
void addInventory(int numItems);
private:
int quantityRemaining;
};
InventoryTag::InventoryTag() {
quantityRemaining = 0;
}
int InventoryTag::getQuantityRemaining() const {
return quantityRemaining;
}
void InventoryTag::addInventory(int numItems) {
if (numItems > 10) {
quantityRemaining = quantityRemaining + numItems;
}
}
int main() {
InventoryTag redSweater;
int sweaterShipment = 0;
int sweaterInventoryBefore = 0;
sweaterInventoryBefore = redSweater.getQuantityRemaining();
sweaterShipment = 25;
cout << "Beginning tests." << endl;
// FIXME add unit test for addInventory
/* Your solution goes here */
cout << "Tests complete." << endl;
return 0;
}</span>
This is something you would do through trial and error. At least, that's the approach I took. I'm not sure if there is any algorithm to solve. The solution I got is shown in the attached image below. There are probably other solutions possible. The trick is to keep each number separate but not too far away so that the other numbers to be filled in later don't get too crowded to their neighbor.
Side note: any mirror copy of what I posted would work as well since you can flip the page around and it's effectively the same solution.
Answer:
y axis
Step-by-step explanation:
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>
Answer:
x=7
Step-by-step explanation:
3(x+1)+5(x+1)=64
(3+5)(x+1)=64
8(x+1)=64
8x+8=64
8x=56
x=7
