To expand the given expression we proceed as follows:
(6x²-2x-6)(8x²+7x+8)
=6x²(8x²+7x+8)-2x(8x²+7x+8)-6(8x²+7x+8)
=48x⁴+42x³+48x²-16x³-14x²-16x-48x²-42x-48
putting like terms together:
48x⁴+(42x³-16x³)+(48x²-48x²)+(-16x-42x)-48
=48x⁴+26x³+0x²-58x-48
hence the answer is:
48x⁴+26x³-58x-48
11 times with a remainder of 5
Answer:
Yes
Step-by-step explanation:
For number 3 you answered them all correctly as well as number 4 and 5. Don't hesitate asking questions like these :D
x⁸ + 48x⁷ + 1008x⁶ + 12096x⁵ + 90720x⁴ + 435456x³ + 1306368x² + 2239488x + 1679616
<u>Explanation:</u>
We know
(x+y)ⁿ = ∑ ⁿCₐxⁿ⁻ᵃyᵃ
and ⁿCₐ = n! / ( a! ) . ( n-a )!
So,
(x+6)⁸ = ⁸C₀x⁸ + ⁸C₁(x)⁸⁻¹(6)¹ + ⁸C₂(x)⁸⁻²(6)² + ⁸C₃(x)⁸⁻³(6)³ + .......+ ⁸C₈(x)⁸⁻⁸(6)⁸
= ₓ⁸ + 8x⁷ₓ 6 + 28x⁶ₓ 36 + 56x⁵ₓ 216 + 70x⁴ₓ 1296 + 56x³ₓ 7776 + 28x²ₓ 46656 + 8x . 279936 + 1679616
= x⁸ + 48x⁷ + 1008x⁶ + 12096x⁵ + 90720x⁴ + 435456x³ + 1306368x² + 2239488x + 1679616
Thus, the expansion of ( x+6)⁸ using binomial theorm is
x⁸ + 48x⁷ + 1008x⁶ + 12096x⁵ + 90720x⁴ + 435456x³ + 1306368x² + 2239488x + 1679616
Answer: She can buy up to 2 bags of nuts
Step-by-step explanation:
Hi, to answer this question we have to write an inequality:
The product of the number of bags of nuts bought (n) and the price per bag (5.70); plus the price of one box of cookies (4.25) must be less or equal to Annie's money (18).
5.70n +4.25 ≤18
Solving for x:
5.70n ≤18 -4.25
5.70n ≤13.75
n ≤13.75/5.70
n ≤2.41
n ≤2 (rounded)
She can buy up to 2 bags of nuts
