In order to solve by elimination, let's multiply the first equation by -2. This way, when we add the equations, the variable y will be canceled out:

Now, adding the equations, we have:

Now, calculating the value of y, we have:

Therefore the solution is (3, 0).
Opposite angles of an inscribed quadrilateral are supplementary, so it is true that ...
... B. m∠A + m∠C = m∠B + m∠D . . . . . = 180°
First of all, let's recall the area of a triangle, knowing its base (b) and height (h):

The exercise is showing you that, if you inscribe a polynomial with more and more side, the area of the polynomial will approximate the area of the circle better and better (you can see youself that the polygon is "filling" the circle more and more as the number of sides increase).
Now, the second column tells you the area of each of the triangles the polygon is split into. So, we can see that the first polygon is split into 3 triangles, each of them having base 1.73 and height 0.5.
So, the area of each triangle is

There are three of these triangles, so the area of the whole polygon is

In the second case, you have six triangles, each with base 1 and height 0.87. So, the whole area is

Finally, in the last case you have 8 triangles, each with base 0.77 and height 0.92. So, the whole area is

It's a <span>rhombus.
1. D
2. A
3. D
4. B
5. C</span>
To answer this question, you need to know that a quarter is worth $0.25 and dimes worth $0.10. In this case, the total number of money that made from quarters and dimes is $2.80
With Q=quarter and D=dime, then the equation should be:
2.80= 0.25 Q + 0.10 D
The condition that needs to be fulfilled in this case is that the quarter should be 8 more than dimes. Then you need to find out the highest possible number of the quarter and lowest possible number of the dimes.
If you divide 2.8 with 0.25 you will found it will be 11.2 but 11 quarter will result with 0.05 value which you cannot remove, so the highest possible of the quarter should be 10.
If the highest possible of the quarter is 10, then the lowest possible of dimes should be:
2.8= 0.25(10) + 0.10D
0.10D= 0.3
D=3
Since 10-3 is 7, then it is not possible for the quarter to be 8 more than the dimes.