Btw, I'm new here. What is the constant term in the expression 6x3y + 8x2 + 5x + 7? (Input a numeric value only.)

I need an answer not a link that don't work!

Mila [183]

Answer:

B

Step-by-step explanation:

6 0

2 years ago

GIVING BRAINLIEST FOR BEST ANSWER!

Stels [109]

Answer: The answer is 11 1/24 or in decimal form 11.04

Step-by-step explanation:

7 0

2 years ago

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Factor this polynomial completely.<br> x^2-8x+12

Sphinxa [80]

Answer:

(x - 2) (x - 6)

Step-by-step explanation:

Factor the following:

x^2 - 8 x + 12

The factors of 12 that sum to -8 are -2 and -6. So, x^2 - 8 x + 12 = (x - 2) (x - 6):

Answer: (x - 2) (x - 6)

7 0

3 years ago

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Prove that if x is an positive real number such that x + x^-1 is an integer, then x^3 + x^-3 is an integer as well.

Shkiper50 [21]

Answer:

By closure property of multiplication and addition of integers,

If $x + \dfrac{1}{x}$ is an integer

∴ $\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer

Step-by-step explanation:

The given expression for the positive integer is x + x⁻¹

The given expression can be written as follows;

$x + \dfrac{1}{x}$

By finding the given expression raised to the power 3, sing Wolfram Alpha online, we we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot x + \dfrac{3}{x}$

By simplification of the cube of the given integer expressions, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 = x^3 + \dfrac{1}{x^3} +3\cdot \left (x + \dfrac{1}{x} \right )$

Therefore, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= x^3 + \dfrac{1}{x^3}$

By rearranging, we get;

$x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )$

Given that $x + \dfrac{1}{x}$ is an integer, from the closure property, the product of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3$ is an integer and $3\cdot \left (x + \dfrac{1}{x} \right )$ is also an integer

Similarly the sum of two integers is always an integer, we have;

$\left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

$\therefore x^3 + \dfrac{1}{x^3} = \left ( x + \dfrac{1}{x} \right) ^3 - 3\cdot \left (x + \dfrac{1}{x} \right )= \left ( x + \dfrac{1}{x} \right) ^3 + \left(- 3\cdot \left (x + \dfrac{1}{x} \right ) \right )$ is an integer

From which we have;

$x^3 + \dfrac{1}{x^3}$ is an integer.

4 0

2 years ago

Use the function below to find f(1). f9t0=4

ExtremeBDS [4]

The value of the F(1) is 1/2 option (C) is correct after plugging the value of t = 1 in the provided function.

<h3>What is a function?</h3>

It is defined as a special type of relationship, and they have a predefined domain and range according to the function every value in the domain is related to exactly one value in the range.

The question is incomplete.

The complete question is attached in the picture please refer to the picture.

We have a function:

$\rm F(t) = 4\dfrac{1}{2^{3t}}$