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Lyrx [107]
3 years ago
9

Simplify. 12÷(2+4) Please help

Mathematics
2 answers:
Georgia [21]3 years ago
8 0
Step^1:  \frac{12}{(2 +4)} \\ Step^2: 2 + 4 = 6 \\ Step^3:  \frac{12}{(6)} = 2 \\ Answer: 2 \\ \\ \\ good \\ luck \\ on \\ your \\ assignment \\ \\ \\ enjoy \\ your \\ day \\ \\ MeIsKaitlyn :)

We found our answer by doing PEMDAS  (Parentheses \\ equation \\ multiplication \\ division \\ addition \\ subtraction)

So, in this problem we were doing  Parentheses, Addition, Division


timurjin [86]3 years ago
6 0
12 / (2 + 4) =
12 / 6 =
2 <==
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• How do you use the distance formula and slope formula to prove properties of polygons?
patriot [66]

Answer:

distance formula= \sqrt{x} (x_{2} -x_{1} )^{2} + (y_{2} -y_{1} )^{2} = find the length(units) using coordinates.

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When working with polygons the main properties which are important are:

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The length of the sides of the shape.

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6 0
3 years ago
Can someone explain this to me please
IrinaVladis [17]

Answer:

c. 36·x

Step-by-step explanation:

Part A

The details of the circle are;

The area of the circle, A = 12·π cm²

The diameter of the circle, d = \overline {AB}

Given that \overline {AB} is the diameter of the circle, we have;

The length of the arc AB = Half the the length of the circumference of the circle

Therefore, we have;

A = 12·π = π·d²/4 = π·\overline {AB}²/4

Therefore;

12 = \overline {AB}²/4

4 × 12 = \overline {AB}²

\overline {AB}² = 48

\overline {AB} = √48 = 4·√3

\overline {AB} = 4·√3

The circumference of the circle, C = π·d = π·\overline {AB}

Arc AB = Half the the length of the circumference of the circle = C/2

Arc AB = C/2 = π·\overline {AB}/2

\overline {AB} = 4·√3

∴ C/2 = π·4·√3/2 = 2·√3·π

The length of arc AB = 2·√3·π cm

Part B

The given parameters are;

The length of \overline {OF} = The length of \overline {FB}

Angle D = angle B

The radius of the circle = 6·x

The measure of arc EF = 60°

The required information = The perimeter of triangle DOB

We have;

Given that the base angles of the triangles DOB are equal, we have that ΔDOB is an isosceles triangle, therefore;

The length of \overline {OD} = The length of \overline {OB}

The length of \overline {OB} = \overline {OF} + \overline {FB} = \overline {OF} + \overline {OF} = 2 × \overline {OF}

∴ The length of \overline {OD} = 2 × \overline {OF} = The length of \overline {OB}

Given that arc EF = 60°, and the point 'O' is the center of the circle, we have;

∠EOF = The measure of arc EF = 60° = ∠DOB

Therefore, in ΔDOB, we have;

∠D + ∠B = 180° - ∠DOB = 180° - 60° = 120°

∵ ∠D = ∠B, we have;

∠D + ∠B = ∠D + ∠D = 2 × ∠D = 120°

∠D = ∠B = 120°/2 = 60°

All three interior angles of ΔDOB = 60°

∴ ΔDOB is an equilateral triangle and all sides of ΔDOB are equal

Therefore;

The length of \overline {OD} = The length of \overline {OB} = The length of \overline {DB}  = 2 × \overline {OF}

The perimeter of ΔDOB = The length of \overline {OD} + The length of \overline {OB} + The length of \overline {DB} = 2 × \overline {OF} + 2 × \overline {OF} + 2 × \overline {OF} = 6 × \overline {OF}

∴ The perimeter of ΔDOB = 6 × \overline {OF}

The radius of the circle = \overline {OF} = 6·x

∴ The perimeter of ΔDOB = 6 × 6·x = 36·x

3 0
3 years ago
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