Eliminate x's by adding te 2 equations
add them
-2x+15y=24
<u>2x+9y=24 +</u>
0x+24y=48
24y=48
divide both sides by 24
y=2
sub back
2x+9y=24
2x+9(2)=24
2x+18=24
minus 18 both sides
2x=6
divide 2
x=3
x=3
y=2
(x,y)
(3,2)
Answer:
a
Step-by-step explanation:
Explanation:
It helps to understand the process of multiplying the binomials. Consider the simple case ...
(x +a)(x +b)
The product is ...
(x +a)(x +b) = x² +(a+b)x + ab
If the <em>constant</em> term (ab) is <em>negative</em>, the signs of (a) and (b) are <em>different</em>.
If the constant term (ab) is <em>positive</em>, the signs of (a) and (b) will both match the sign of the coefficient of the linear term (a+b).
___
Of course, the sum (a+b) will have the sign of the (a) or (b) value with the largest magnitude, so when the signs of (a) and (b) are different, the factor with the largest magnitude will have the sign of (a+b), the x-coefficient.
<u>Example</u>:
x² -x -6
-6 tells you the factors will have different signs. -x tells you the one with the largest magnitude will be negative.
-6 = -6×1 = -3×2 = ... (other factor pairs have a negative factor with a smaller magnitude)
The sums of these factor pairs are -5 and -1. We want the factor pair that has a sum of -1, the coefficient of x in the trinomial.
x² -x -6 = (x -3)(x +2)
Answer:
2
Step-by-step explanation:
Brainliest?!?!?!?
Answer:
Rhombus
Step-by-step explanation:
The given points are A(−5, 6), B(−1, 8), C(3, 6), D(−1, 4).
We use the distance formula to find the length of AB.
![|AB|=\sqrt{(-1--5)^2+(8-6)^2}](https://tex.z-dn.net/?f=%7CAB%7C%3D%5Csqrt%7B%28-1--5%29%5E2%2B%288-6%29%5E2%7D)
![|AB|=\sqrt{16+4}](https://tex.z-dn.net/?f=%7CAB%7C%3D%5Csqrt%7B16%2B4%7D)
![|AB|=\sqrt{20}](https://tex.z-dn.net/?f=%7CAB%7C%3D%5Csqrt%7B20%7D)
The length of AD is
![|AD|=\sqrt{(-1--5)^2+(6-4)^2}](https://tex.z-dn.net/?f=%7CAD%7C%3D%5Csqrt%7B%28-1--5%29%5E2%2B%286-4%29%5E2%7D)
![|AD|=\sqrt{16+4}](https://tex.z-dn.net/?f=%7CAD%7C%3D%5Csqrt%7B16%2B4%7D)
![|AD|=\sqrt{20}](https://tex.z-dn.net/?f=%7CAD%7C%3D%5Csqrt%7B20%7D)
The length of BC is:
![|BC|=\sqrt{(-1-3)^2+(8-6)^2}](https://tex.z-dn.net/?f=%7CBC%7C%3D%5Csqrt%7B%28-1-3%29%5E2%2B%288-6%29%5E2%7D)
![|BC|=\sqrt{16+4}](https://tex.z-dn.net/?f=%7CBC%7C%3D%5Csqrt%7B16%2B4%7D)
![|BC|=\sqrt{20}](https://tex.z-dn.net/?f=%7CBC%7C%3D%5Csqrt%7B20%7D)
The length of CD is
![|CD|=\sqrt{(-1-3)^2+(6-4)^2}](https://tex.z-dn.net/?f=%7CCD%7C%3D%5Csqrt%7B%28-1-3%29%5E2%2B%286-4%29%5E2%7D)
![|CD|=\sqrt{16+4}](https://tex.z-dn.net/?f=%7CCD%7C%3D%5Csqrt%7B16%2B4%7D)
![|CD|=\sqrt{20}](https://tex.z-dn.net/?f=%7CCD%7C%3D%5Csqrt%7B20%7D)
Since all sides are congruent the quadrilateral could be a rhombus or a square.
Slope of AB![=\frac{8-6}{-1--5}=\frac{1}{2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B8-6%7D%7B-1--5%7D%3D%5Cfrac%7B1%7D%7B2%7D)
Slope of BC ![=\frac{8-6}{-1-3}=-\frac{1}{2}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B8-6%7D%7B-1-3%7D%3D-%5Cfrac%7B1%7D%7B2%7D)
Since the slopes of the adjacent sides are not negative reciprocals of each other, the quadrilateral cannot be a square. It is a rhombus