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Digiron [165]
3 years ago
10

\angle x∠xangle, x and \angle y∠yangle, y are supplementary angles. \angle y∠yangle, y measures 156^\circ156 ∘ 156, degrees. Wha

t is the measure of \angle x∠xangle, x?
Mathematics
1 answer:
dem82 [27]3 years ago
6 0

The measure of x is  \angle x=24^{\circ}

Explanation:

It is given that $\angle x$ and $\angle y$ are supplementary angles.

The two angles are said to be supplementary if their angles add up to 180°

Thus, we have,

\angle x+\angle y=180^{\circ}

Also, it is given that \angle y=156^{\circ}

Substituting the value in the expression \angle x+\angle y=180^{\circ}, we have,

\angle x+$156^{\circ}$=180^{\circ}

Subtracting both sides by $156^{\circ}$, we get,

\angle x+$156^{\circ}$-156^{\circ}=180^{\circ}-156^{\circ}

                 \angle x=24^{\circ}

Thus, the measure of x is \angle x=24^{\circ}

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Helen [10]

Partial Answer:

For #10 the solutions are 2 and 5

Step-by-step explanation:

Solutions for an equation can be x-intercepts, or where it touches the x or horizontal line. The equation in #10 touches the x line at 2 and 5.

5 0
2 years ago
The value of x is Helllpp
rewona [7]

a.84

clickkkkk

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6 0
3 years ago
Read 2 more answers
For the graph shown to the right, find (a) AB to the nearest tenth and (b) the coordinates of the midpoint of AB.
Shkiper50 [21]
AB =

\sqrt{( {4 -  (- 2))}^{2}  +  {( - 5 + 3)}^{2} }  =  \sqrt{40}  = 6.3
The midpoint os segment AB =

(( \frac{ - 2 + 4}{2} ) = the \: x \: coordinate \: of \: the \: midpoint \\  \\ ( \frac{ - 3 - 5}{2})) = the \: y \: coordinate \: of \: the \: midpoint
The midpoint is:

(1,-4)
8 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
3 years ago
Plz help ASAP, thanks​
astraxan [27]

Answer:

12

Step-by-step explanation:

8·15·h=1440

120·h=1440

\frac{120h}{120}=\frac{1440}{120} (Divide both sides by 120)

h=12

<em><u>h stands for height</u></em>

6 0
2 years ago
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