<span>a) We know that the correct answer will be the square root of 256 since the competition area is a square with an area of 256 meters. And since 10^2 = 100 which is less than 256, the answer has to be greater than 10. And since 20^2 = 400 which is greater than 256, the answer also has to be less than 20. Therefore the answer has to be between 10 and 20.
b) The last digit has to be either a 4 or a 6. The units digit is the only digit that will contribute to the units digit of the square. And 0^2 = 0, 1^2 = 1, 2^2 = 4, 3^2 = 9, 4^2 = 16, 5^2 = 25, 6^2 = 36, 7^2 = 49, 8^2 = 64, 9^2 = 81. Of the 10 possible digits, only the values 4 and 6 have a square that has an units digit of 6.
c) The square root of 256 based up (a) and (b) above has to be either 14, or 16. So the dimensions are either 14x14 meters or 16x16 meters.</span>
Answer:
89/16
Step-by-step explanation:
You have to use the formule: y2-y1/x2-x1
-2-(-2)/4-(-3)=0/7
(8x+3)(x-2)
Explanation
8x^2-13x-6
X(8x+3)-2(8x+3) factor buck greatest common factor
(8x+3)(x-2)
3x + y = 3
7x + 2y = 1
First isolate one of the variables (x or y) in one of the equations.
Isolate "y" in the first equation(because it is the easiest to isolate) and substitute it into the second equation.
3x + y = 3 Subtract 3x on both sides
3x - 3x + y = 3 - 3x
y = 3 - 3x
7x + 2y = 1
7x + 2(3 - 3x) = 1 [since y = 3 - 3x, you can substitute (3-3x) for "y"]
Multiply/distribute 2 into (3 - 3x)
7x + (3(2) - 3x(2)) = 1
7x + 6 - 6x = 1
x + 6 = 1 Subtract 6 on both sides
x = -5
Now that you know "x", substitute it into one of the equations (I will do both)
3x + y = 3
3(-5) + y = 3 [since x = -5, you can plug in -5 for "x"]
-15 + y = 3 Add 15 on both sides
y = 18
7x + 2y = 1
7(-5) + 2y = 1
-35 + 2y = 1 Add 35 on both sides
2y = 36 Divide 2 on both sides
y = 18
x = -5, y = 18 or (-5, 18)
