I believe the answer is b and e
The first reaction represented is a decomposition reaction due to the fact the bonded pair are being split apart.
The second equation is a single displacement because the B and C have switched places, which is the only change making it a single displacement. If there had been another bond and A had also moved, it would be double.
<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>
For -4+3,
1. start with zero
2. move to left (-x axis) for 4 units (-4)
3. then move to right (+x axis) for 3 units (+3)
For 3+(-4)
1. Start with zero
2. Move the point to right side by 3 units (+3)
3. then move the point to the left by 4 units (-4)
For both you'll stop at -1.
F(x) = -x² + 4
f(3) = -3² + 4
f(3) = -9 +4
f(3) = -5
