Answer:
1, 9, 36, 84, 126, 126, 84, 36, 9, 1.
Step-by-step explanation:
Using the binomial theorem:
(x + y)^9 = x^9 + 9C1 x^8 y + 9C2 x^7y^2 + 9C3x^6y^3 + 9C4x^5y^4 + 9C5x^4y^5 +
9C6x^3y^6 + 9C7x^2y^7 + 9C8xy^8 + y^9
The coefficients are
1, 9, 36, 84, 126, 126, 84, 36, 9, 1.
Step 1 : Setting up the problem
Write the coefficients of the dividend in the same order. For missing terms, enter the co-efficient as zero. Set the divisor equal to zero and use that number in the division box.
The problem now looks as follows:
-1 | 12 5 3 0 -5
Step 2 : Bring down the first co-efficient and write it in the bottom row.
-1 | 12 5 3 0 -5
______________________ 12
Step 3 : Multiply the first coefficient with the divisor and enter the value below
the next co-efficient. Add the two and write the value in the bottom row.
-1 | 12 5 3 0 -5
_____-12_______________ 12 -7
Step 4 : Repeat Step 3 for rest of the coefficients as well:
-1 | 12 5 3 0 -5
____________7 ___________ 12 -7 10
-1 | 12 5 3 0 -5 ______________ -10______ 12 -7 10 -10
-1 | 12 5 3 0 -5 ____________________ 10_ 12 -7 10 -10 5
The last row now represents the quotient coefficients and the remainder. Co-efficients of Quotient are written one power less than their original power and the remainder is written as a fraction.
Answer :12x^3-7x^2+10x-10+5/(x+1) where the last term denotes the remainder and the rest is the quotient.
Answer: 21.3 miles
Step-by-step explanation:
Answer:
The multiplicative inverse of y is 
.
Step-by-step explanation:
Multiplicative inverse implies a value that you would multiply with the given value so as to give 1.
Let the multiplicative inverse of y be represented by a, so that;
a x y = 1
ay = 1
divide both sides by y to have,
a =
Thus, the multiplicative inverse of y is .
This is applicable to any given value. Examples:
i. Given 2, the multiplicative inverse is 
.
⇒ 2 x = 1
ii. What is the multiplicative inverse of 5?
The required answer is 
.
Answer:
A is a function; B, C and D are not.
Step-by-step explanation:
In other words: Identify the sole function among these relationships.
A function maps any input onto exactly one output.
If a relationship maps any input onto more than one output, it is not a function.
Thus, we eliminate B, C and D. In B, for example, we have the inputs {1, 2, 3}, where the '1' has two y-values associated with it.
On the other hand, A has the domain {-1, 0, 1, 2}, and all four inputs have exactly one associated output.
