Answer:
$160
Step-by-step explanation:
Part A-
Step 1: 6(2k+5) - 20
Step 2: 6(2k) + 6(5) - 20
Step 3: 12k + 30 - 20
<h2><u>
12k + 10</u></h2><h2><u>
</u></h2><h3>Part B-</h3><h3>12(12.50) + 10</h3><h3>150 + 10</h3><h2><u>$160</u></h2>
Answer:
we have:
8x³ + mx² - 6x + n
= 8x³ - 8x² + (m + 8)x²- (m + 8)x + (m + 2)x - (m + 2) + m + 2+ n
= 8x²(x - 1) + (m + 8)x(x - 1) + (m + 2)(x - 1) + (m + n + 2)
= (x - 1)[8x² + (m + 8)x + m + 2] + (m + n + 2)
because the remainder if divided by (x-1) is 2
=> m + n + 2 = 2
⇔ m + n = 0 (1)
we also have:
8x³ + mx² - 6x + n
= 8x³ - 12x² + (m + 12)x² - 3/2.x.(m + 12) + ( 12 + 3/2.m)x - (9/4.m + 18) + n +9/4m + 18
= 4x²(2x - 3) + 1/2.(m + 12)x(2x - 3) + (3/2m + 12).1/2.(2x - 3) + 9/4m + n + 18
= (2x - 3)(4x² + (m + 12)/2.x + 3/4m + 6) + 9/4m + n + 18
because the remainder if divided by (2x - 3) is 8
=> 9/4m + n + 18 = 8
⇔ 9/4m + n = -10 (2)
from (1) and (2), we have:
m + n = 0
9/4m + n = -10
=> m = -8
n = 8
Step-by-step explanation:
Answer:
D) -4 (with imaginary roots ±2i)
Step-by-step explanation:
Since you cannot take the square root of a negative number to produce a real result, the only option that has two imaginary roots is D) -4, where the two imaginary roots are 2i and -2i.
(-2, -1)
(3, 2)
formula:
y2 - y1
———-
x2 - x1
2 - - 1 3
——- = —
3 - - 2 5
