In each case, you can use the second equation to create an expression for y that will substitute into the first equation. Then you can write the result in standard form and use any of several means to find the number of solutions.
System A
x² + (-x/2)² = 17
x² = 17/(5/4) = 13.6
x = ±√13.6 . . . . 2 real solutions
System B
-6x +5 = x² -7x +10
x² -x +5 = 0
The discriminant is ...
D = (-1)²-4(1)(5) = -20 . . . . 0 real solutions
System C
y = 8x +17 = -2x² +9
2x² +8x +8 = 0
2(x+2)² = 0
x = -2 . . . . 1 real solution
Answer:
B
Step-by-step explanation:
One good way to look at this is to graph both polynomials, as shown in the picture. A tip to help graph is to factor it out and work from there. For example, in x²+14x+48, we can gather that (x+6)(x+8) is the same thing, and it is easier to then graph it. Similarly, for x²+12x+36, we can factor it out as (x+6)² .
When x²+12x+36 approaches 6, it is getting really close to 0, but it stays positive. When x²+14x+48 approaches 6 from the negative side, it is also getting close to 0, but it's negative. When x²+14x+48 approaches 6 from the positive side, it is positive.
Therefore, on the negative side, there is one positive and one negative (dividing a negative by a positive is negative, and a positive by a negative is also negative) , and on the positive side, there are two positives, forming one answer.The answer is therefore B
1) is false, it is always a rhombus.
2) is true, a rectangle with all sides equal would be a square.
3) is false, a kite's side lengths must be 2 pairs of unequal lengths.
4) is true, a square is a rhombus that has 90-degree angles.
5) is true, a kite must have 2 pairs of unequal lengths, but a trapezoid cannot have exactly 2 sides of equal lengths.