Identify the equation of the circle that has its center at (9, 12) and passes through the origin.
1 answer:
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Answer:
see explanation
Step-by-step explanation:
Where the line crosses the x- axis the y- coordinate is zero.
Substitute y = 0 into the equation and solve for x
x + 2 = 0 ( multiply through by 4 to clear the fraction )
x + 8 = 0 ( subtract 8 from both sides )
x = - 8 ← x- intercept ⇒ (- 8, 0 )
Where the line crosses the y- axis the x- coordinate is zero.
Substitute x = 0 into the equation and solve for y
y = 0 + 2, that is
y = 2 ← y- intercept ⇒ (0, 2 )
Answer:
Step-by-step explanation:
(e) (x-8)(x-9)
Multiplying two binomials.
x(x-9)-8(x-9)
x² -9x - 8x + 72
x² - 17x + 72
(f) 6x⁴ - 12x
Common factoring.
6x(x³ - 2)
(g) x² + 8x + 12
Factoring trinomial of the form ax² + bx + c.
x² +2x + 6x + 12
x(x + 2) + 6(x + 2)
(x + 2) (x + 6)
(h) (x + 7)(x + 5)
Multiplying two binomials.
x(x + 5)+7(x + 5)
x² + 5x + 7x + 35
x² + 12x + 35
(i) x² - 36
Difference of two squares.
a² - b² = (a + b)(a - b)
x² - 6²
a = x, b = 6
(x + 6)(x - 6)