<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>
b.an object at rest has an instantaneous acceleration of zero
Answer: A
Step-by-step explanation:
Question is not well presented
The parabola y=x² is scaled vertically by a factor of 1/10.
What is the equation of the new parabola?
Answer:
The equation of the new parabola is 0.1x²
Step-by-step explanation:
Given
Parabola: y = x²
Scale = 1/10 = 0.1
The interpretation of this question is that; there's a need to scale the graph in ratio 1:10.
I.e; 1 unit on the parabola is being represented by 10 unit on the scale
So, x (on the new parabola) = 1/10 of old x.
So, the new equation of the parabola = 1/10x²
New equation = 0.1x²
Hence, the equation of the new parabola is calculated as 0.1x²