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Serjik [45]
1 year ago
10

Hurry show your work

Mathematics
1 answer:
Nikolay [14]1 year ago
7 0

Answer: \huge\boxed{\boxed{15}}

Step-by-step explanation:

<u>Given expression</u>

\large\boxed{\frac{12[30 - (9+4^2)]}{|10|-|-6| } }

<u>Simplify the exponents</u>

\large\boxed{=\frac{12[30 - (9+16)]}{|10|-|-6| } }

Simplify values in the parenthesis

\large\boxed{=\frac{12[30 - 25]}{|10|-|-6| } }

\large\boxed{=\frac{12[5]}{|10|-|-6| } }

<u>Simplify absolute values (all positive)</u>

\large\boxed{=\frac{12[5]}{10-6 } }

\large\boxed{=\frac{12[5]}{4 } }

<u>Simplify by division</u>

\large\boxed{=3~[5]}

<u>Simplify by multiplication</u>

\huge\boxed{\boxed{=15}}

Hope this helps!! :)

Please let me know if you have any questions

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Suppose f(x) = 2x-1 and g (x) = x + 1/a. Which value(s) of a would make the composite functions commutative? A. 0 B. 2 C. any va
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2

Step-by-step explanation:

A commutative function means that when you insert one function in space of x in the other function, they will equal x. The equation is f(g(x)) = g(f(x)) = x

So, 2(\frac{x+1}{a}) - 1 = \frac{(2x-1)+1}{a}

If you multiply both sides by <em>a</em>, you get   2<em>a</em>(\frac{x+1}{a}) - a = (2x-1)+1

Simplify it                                                   2<em>a</em>(\frac{x+1}{a}) - <em>a</em> = 2x

Add <em>a</em> to both sides                                    2<em>a</em>(\frac{x+1}{a})  = 2x +<em>a</em>

The two <em>a</em>s on the left cancel out                   2(x+1)=2x+<em>a</em>

Distribute the 2                                               2x+2=2x+<em>a</em>

Then subtract 2x from both sides                        2 = <em>a</em>

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Answer:

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Step-by-step explanation:

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Now to find the area of the triangle.

The triangle is an isosceles triangle That means two of its sides are equal. They are equal to the radius of the circle, which is 24.

the small angles are equal to

2x + 120 = 180    Subtract 120 from both sides

2x = 60               Divide by 2

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The height of the triangle is derived from sin(30) = opposite / hypotenuse

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