56 = 2 x 2 x 2 x 732 = 2 x 2 x 2 x 2 x 2
GCF = 2 x 2 x 2 =8
Rewrite 56+32 as the product of the GCF and a sum:
56 + 32 = 8 (7+4)
(1) y² + x² = 53
(2) y - x = 5 ⇒ y = x + 5
subtitute (2) to (1)
(x + 5)² + x² = 53 |use (a + b)² = a² + 2ab + b²
x² + 2x·5 + 5² + x² = 53
2x² + 10x + 25 = 53 |subtract 53 from both sides
2x² + 10x - 28 =0 |divide both sides by 2
x² + 5x - 14 = 0
x² - 2x+ 7x - 14 = 0
x(x - 2) + 7(x - 2) = 0
(x - 2)(x + 7) = 0 ⇔ x - 2 = 0 or x + 7 = 0 ⇔ x = 2 or x = -7
subtitute the values of y to (2)
for x = 2, y = 5 + 2 = 7
for x = -7, y = 5 + (-7) = 5 - 2 = 3
Answer: x = 2 and y = 7 or x = -7 and y = 3
Answer:

Step-by-step explanation:
I suggest looking lessons on it online if you don't understand... It's hard to do your work especially if you don't understand it. Or if your online, go to past lessons or something.
Answer:
2884.8 millimeters squared
Step-by-step explanation:
Given:
The radius of the disc is 35 millimeters, so the area of it is:
π
= 3.14*
= 3846.5
Then, we find out the area of the circular hold cut out of the bigger one, its radius is a haft of the radius of the bigger circle = 35/2 = 17.5
π
= 3.14*
=961.6
=> the area of the pendant = 3846.5 - 961.6 =2884.8 millimeters squared