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2 years ago

Express 2.41 x 10^4 in a standard form

Mathematics

1 answer:

Dmitrij [34]2 years ago

6 0

Step-by-step explanation:

$2.41 \times {10}^{4} \\ = 2.41 \times 10000 \\ = 24100 \\$

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How do you factor 6x^3 +5x^2-4x

BabaBlast [244]

You can take out an x
X(6x^2+5x-4)
Then factor it
X(3x+4)(2x-1)

6 0

2 years ago

Tim needs 3 pounds of vegetables for a vegetable soup recipe.

klio [65]

Answer:

25/12 - pounds more vegetables

Step-by-step explanation:

first we need to put these variables into like terms

36/12 - 3

8/12 - 2/3

3/12 - 1/4

add the last two up and subtract from the first and you get how much you need left over

8 0

2 years ago

For the functions f(x) = 5x − 3 and g(x) = x + 4, which composition produces the greatest output?

Zina [86]

You could actually find the compositions and thus have something to compare. You haven't shared the list of possible answer choices.

(f+g)(x) = 5x - 3 + x + 4 = 6x + 1

(f-g)(x) = 5x - 3 - x - 4 = 4x - 7

(f*g)(x) = (5x-3)((x+4) = 5x^2 + 20x - 3x - 12 = 5x^2 + 17x - 12

There are also the quotient (f/g)(x) and the compositions f(g(x)) and g(f(x)).

WRite them out.

Then you could arbitrarily select x values, such as 2, 10, etc., subst. them into each composition and determine which output is greatest.

7 0

3 years ago

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

Give the meaning of the equation: 6x + 4y = 2. 3x + 2y = 1

seraphim [82]

6x+4y=2
3x+2y=1

If you multiply by 2 the second statement you'll get the first one which means they belong to the same graph (same line, lines are coincident). The system is consistent and dependent.

The answer is: C. coincident

4 0

2 years ago

Read 2 more answers