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3 years ago

Which numbers are the extremes of the proportion 3:4=6:8

Mathematics

1 answer:

yaroslaw [1]3 years ago

8 0

3×2/4×2

2/2

hope this helps

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How many hours was the trip?

Sidana [21]

Answer:

30 hrs here ya go !!!!!!!!!!!!!!!

7 0

3 years ago

I will give brainlest and 75 points

Lelechka [254]

Answer:

Step-by-step explanation

3.2x+5.7=2.5x. X=-8.142857

10.1x-1.6x+44=-7 x=-6

7.9x+x+4=1.1x-16 x=-2.564103

1/4x+5/2x-2=4-1/4x x=2

5 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

3 years ago

Please Help ASAP!!!!

KIM [24]

Answer:

$m\angle M = 98\degree$

Step-by-step explanation:

As we know that:

Sum of the measure of all interior angles of a triangle is always 180°.

$\therefore (3x-16)\degree+(x+6)\degree+ x\degree = 180\degree\\\therefore (5x-10)\degree = 180\degree\\5x-10 = 180\\5x = 180 + 10\\5x = 190\\x= \frac{190}{5}\\x= 38\\\because m\angle M = (3x-16)\degree\\\therefore m\angle M = (3\times 38 -16)\degree\\\therefore m\angle M = (3\times 38 -16)\degree\\\therefore m\angle M = (114 -16)\degree\\\therefore m\angle M = 98\degree$

3 0

2 years ago

A beach has to enclose a rectangular area, because some endangered species are nesting there. They have 200 feet of rope to rope

kkurt [141]

Area is equal to length times width. The perimeter (the amount of rope) has to equal twice the length added to twice the width so we're left with:
A = l * w
200 = 2l + 2w
solve for either l or w
l = 100 - w
plug into the area equation to get one equation with two variables
A = w(100 - w)
A = -w^2 + 100w
take the derivative
A' = -2w + 100
set the derivative equal to zero
0 = -2w + 100
2w = 100
w = 50
This is the width that maximizes the area
with a width of 50, the length must also be 50 to have a perimeter of 200
therefore, they can rope up to 50 * 50 = 2500 ft^2

6 0

3 years ago