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3 years ago

The slope of the line that passes through the point 2,4 and 3,8 is

Mathematics

2 answers:

Basile [38]3 years ago

8 0

HOPE THIS WILL HELP U:)

Olin [163]3 years ago

4 0

Answer:3

Step-by-step explanation:

8-4/2-3

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2.240 F = d + 2.24 What is the answer

ExtremeBDS [4]

Answer:

F= 0.44642857d+1

Step-by-step explanation:

4 0

2 years ago

An apple producer sells their product in bags that each contain 161616 apples. These apples have a mean weight of 808080 grams a

Dominik [7]

Answer:

Sample mean weight = 808080 grams

Step-by-step explanation:

The number of samples, n = 161616

The mean weight, $\mu = 808080$

Standard deviation, $\sigma = 555$

The sample mean weight, $\mu_{\bar{x}}$ = Mean weight, $\mu$

Since $\mu_{\bar{x}} = \mu$

Sample mean weight, $\mu_{\bar{x}} = 808080$

6 0

3 years ago

W(x) = −3x− 4; Find w(7)

Arturiano [62]

Step-by-step explanation:

Given

w(x) = - 3x - 4

Now

w(7) = - 3 * 7 - 4

= - 21 - 4

= - 25

Hope it will help :)❤

4 0

2 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

Given the definitions of f(x) and g(x) below, find the value of f(g(0)).

spayn [35]

Answer:

f(g(0)) = 45

Step-by-step explanation:

to evaluate f(g(0)) , evaluate g(0) then substitute the value obtained into f(x)

g(0) = - 3(0) - 3 = 0 - 3 = - 3 , then

f(- 3) = 2(- 3)² - 5(- 3) + 12

= 2(9) + 15 + 12

= 18 + 27

= 45

3 0

1 year ago