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riadik2000 [5.3K]
3 years ago
13

To evaluate 4^3/2 find

Mathematics
2 answers:
Vera_Pavlovna [14]3 years ago
8 0

Answer:

32

Step-by-step explanation:

follow the order of operations: PEMDAS

Calculate the exponents

4³

4³=64

=64÷2

Multiply and divide (left to right)

64÷2

=32

Answer

32

hope this helps

Lilit [14]3 years ago
5 0

Answer:

32

Step-by-step explanation:

Order of Operations: BPEMDAS

B - Brackets

P - Parenthesis

M - Multiplication

D - Division

A - Addition

S - Subtraction

Step 1: Write expression

4³/2

Step 2: Exponents

64/2

Step 3: Divide

32

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Step-by-step explanation:

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If a diameter intersects a chord of a circle at a right angle, what conclusion can be made?
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Answer:

The chord is bisected.

Step-by-step explanation:

see the attached figure to better understand the problem

In the circle of the figure

The diameter is the segment DE

The chord is the segment AB

PA=PB=r ----> radius of the circle

Triangles PAC and PBC  are congruent right triangles by SSS

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AC=BC ----> Applying Pythagoras Theorem

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The chord AB is bisected

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(7+5 + 4 . 13 - 2 <br><br> Please simplify this expression with all steps thank you
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Answer:

We apply the BODMAS rule

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3 years ago
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The perimeter of equilateral triangle ABC is 81/3 centimeters, find the length of the radius and apothem.
MAXImum [283]

There is a typo error, the perimeter of equilateral triangle ABC is 81/√3 centimeters.

Answer:

Radius = OB= 27 cm

Apothem = 13.5 cm

A diagram is attached for reference.

Step-by-step explanation:

Given,

The perimeter of equilateral triangle ABC is 81/√3 centimeters.

Substituting this in the formula of perimeter of equilateral triangle =3\times\ side

3\times\ side =[tex]81\sqrt{3}

Side = \frac{81\sqrt{3} }{3} =27\sqrt{3} \ cm

Thus from the diagram , Side AB=BC=AC= 27\sqrt{3} \ cm

We know each angle of an equilateral triangle is 60°.

From the diagram, OB is an angle bisector.

Thus \angle OBC = 30°

Apothem is the line segment from the mid point of any side to the center the equilateral triangle.

Therefore considering ΔOBE, and applying tan function.

tan\theta =\frac{perpendicular}{base} \\tan\theta=\frac{OE}{BE} \\tan\theta=\frac{OE}{\frac{27\sqrt{3}}{2}  } \\tan30\times {\frac{27\sqrt{3} }{2} }= OE\\\frac{1}{\sqrt{3} } \times\frac{27\sqrt{3} }{2} =OE\\

Thus ,apothem  OE= 13.5 cm

Now for radius,

We consider ΔOBE

cos\theta=\frac{base}{hypotenuse} \\cos30= \frac{BE}{OB} \\Cos30 = \frac{\frac{27\sqrt{3} }{2}}{OB}  \\OB= \frac{\frac{27\sqrt{3} }{2}}{cos30} \\OB= \frac{\frac{27\sqrt{3} }{2}}{\frac{\sqrt{3} }{2} } \\OB =27 \ cm

Thus for

Perimeter of equilateral triangle ABC is 81/√3 centimeters,

The radius of equilateral triangle ABC is 27 cm

The apothem of equilateral triangle ABC is 13.5 cm

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3 years ago
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