Do the data in the table represent a direct variation or an inverse variation? Write an equation to model the data in the table.
x 1 3 5 10
y 4 12 20 40
a. direct variation; y=4x
b. direct variation; xy=1/4
c. inverse variation; xy=4
d. inverse variation; xy=1/4
9 answers
20 = 4 * 5
Has anyone found the answer? i cant figure it out :(
It’s a
Do the data in the table represent a direct variation or an inverse variation? Write an equation to model the data in the table.
x 1 3 5 10
y 4 12 20 40
a. direct variation; y=4x
b. direct variation; xy=1/4
c. inverse variation; xy=4
d. inverse variation; xy=1/4
x 1 3 5 10
y 4 12 20 40
a. direct variation; y=4x
b. direct variation; xy=1/4
c. inverse variation; xy=4
d. inverse variation; xy=1/4
The data represents direct variation.
The equation to model the data would be: y = 4x
The equation to model the data would be: y = 4x
Find the LCD of the pair of expressions.
2/7x^10y^4 x 2/14x^8y^7
2/7x^10y^4 x 2/14x^8y^7
The prime factorization of each denominator is:
7x^10y^4 = 7 * x^10 * y^4
14x^8y^7 = 2 * 7 * x^8 * y^7
The LCD includes the highest power of each prime factor that appears in either denominator. Thus, the LCD of 2/7x^10y^4 and 2/14x^8y^7 is:
2 * 7 * x^10 * y^7
Therefore, the LCD is 14x^10y^7.
7x^10y^4 = 7 * x^10 * y^4
14x^8y^7 = 2 * 7 * x^8 * y^7
The LCD includes the highest power of each prime factor that appears in either denominator. Thus, the LCD of 2/7x^10y^4 and 2/14x^8y^7 is:
2 * 7 * x^10 * y^7
Therefore, the LCD is 14x^10y^7.
Use the remainder theorem to determine which of the following binomials are factors of 2x^3+6x^2+8. Select all that apply.
(x-4)
(x-1)
(x+1)
(x+2)
(x+8)
(x-4)
(x-1)
(x+1)
(x+2)
(x+8)
We can use the remainder theorem by dividing the polynomial 2x^3 + 6x^2 + 8 by each of the given binomials and checking if the remainder is zero.
When divided by (x-4), we get:
2x^3 + 6x^2 + 8 = (x-4)(2x^2 + 4x + 10) + 48
Since the remainder is not zero, (x-4) is not a factor of 2x^3 + 6x^2 + 8.
When divided by (x-1), we get:
2x^3 + 6x^2 + 8 = (x-1)(2x^2 + 8x + 8) + 0
Since the remainder is zero, (x-1) is a factor of 2x^3 + 6x^2 + 8.
When divided by (x+1), we get:
2x^3 + 6x^2 + 8 = (x+1)(2x^2 + 4x + 4) + 4
Since the remainder is not zero, (x+1) is not a factor of 2x^3 + 6x^2 + 8.
When divided by (x+2), we get:
2x^3 + 6x^2 + 8 = (x+2)(2x^2 + 2x + 2) + 4
Since the remainder is not zero, (x+2) is not a factor of 2x^3 + 6x^2 + 8.
When divided by (x+8), we get:
2x^3 + 6x^2 + 8 = (x+8)(2x^2 - 10x + 82) - 648
Since the remainder is not zero, (x+8) is not a factor of 2x^3 + 6x^2 + 8.
Therefore, the only binomial that is a factor of 2x^3 + 6x^2 + 8 is (x-1).
When divided by (x-4), we get:
2x^3 + 6x^2 + 8 = (x-4)(2x^2 + 4x + 10) + 48
Since the remainder is not zero, (x-4) is not a factor of 2x^3 + 6x^2 + 8.
When divided by (x-1), we get:
2x^3 + 6x^2 + 8 = (x-1)(2x^2 + 8x + 8) + 0
Since the remainder is zero, (x-1) is a factor of 2x^3 + 6x^2 + 8.
When divided by (x+1), we get:
2x^3 + 6x^2 + 8 = (x+1)(2x^2 + 4x + 4) + 4
Since the remainder is not zero, (x+1) is not a factor of 2x^3 + 6x^2 + 8.
When divided by (x+2), we get:
2x^3 + 6x^2 + 8 = (x+2)(2x^2 + 2x + 2) + 4
Since the remainder is not zero, (x+2) is not a factor of 2x^3 + 6x^2 + 8.
When divided by (x+8), we get:
2x^3 + 6x^2 + 8 = (x+8)(2x^2 - 10x + 82) - 648
Since the remainder is not zero, (x+8) is not a factor of 2x^3 + 6x^2 + 8.
Therefore, the only binomial that is a factor of 2x^3 + 6x^2 + 8 is (x-1).
5 answers