Answer:
Part A : y²(x + 2)(x + 4)
Part B: (x + 4) (x + 4)
Part C: (x + 4) (x - 4)
Step-by-step explanation:
Part A: Factor x²y²+ 6xy²+ 8y²
x²y²+ 6xy²+ 8y²
y² is very common across the quadratic equation , hence
= y² (x² + 6x + 8)
= (y²) (x² + 6x + 8)
= (y²) (x² + 2x +4x + 8)
= (y²) (x² + 2x)+(4x + 8)
= (y²) (x(x + 2)+ 4(x + 2))
= y²(x+2)(x+4)
Part B: Factor x² + 8x + 16
x² + 8x + 16
= x² + 4x + 4x + 16
= (x² + 4x) + (4x + 16)
= x( x + 4) + 4(x + 4)
= (x + 4) (x + 4)
Part C: Factor x² − 16
= x² − 16
= x² + 0x − 16
= x² + 4x - 4x - 16
= (x² + 4x) - (4x - 16)
= x (x + 4) - 4(x + 4)
= (x + 4) (x - 4)
Answer:
3 each
Step-by-step explanation:
36 / 12 = 3
.................................................
Answer:
14
Step-by-step explanation:
Simplify the following:
5 - 2×9 + 9^2/3
Hint: | Evaluate 9^2.
9^2 = 81:
5 - 2×9 + 81/3
Hint: | Reduce 81/3 to lowest terms. Start by finding the GCD of 81 and 3.
The gcd of 81 and 3 is 3, so 81/3 = (3×27)/(3×1) = 3/3×27 = 27:
5 - 2×9 + 27
Hint: | Multiply -2 and 9 together.
-2×9 = -18:
5 + -18 + 27
Hint: | Evaluate 5 + 27 using long addition.
| 1 |
| 2 | 7
+ | | 5
| 3 | 2:
32 - 18
Hint: | Subtract 18 from 32.
| 2 | 12
| 3 | 2
- | 1 | 8
| 1 | 4:
Answer: 14
<span>x1+x2+x3=0 if x1,x2,x3 > -5?
and the solutions are integer
x1= - 4 and x2+x3=4 we count (-4;4;0), (-4;0;4), (-4;3;1), (-4;1;3), (-4; 2;2), (-4;2;2)
x1= -3 and x2+x3=3 we count (-3; 0;3), (-3;3;0), (-3;1;2), (-3;2;1)
x1= -2 and x2+x3=2 we count (-2;0;2). (-2;2;0), (-2;1;1)
x1= -1 and x2+x3=1 we count (-1;0;1), (-1;1;0)
x1=0 and (0;0;0), (0;1;-1), (0;-1;1)
There are 42 solutions
Have fun</span>
