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juin [17]
2 years ago
5

10 points!!! What is sqr root 12x^8/ sqr root 3x^2 in simplest form

Mathematics
2 answers:
marta [7]2 years ago
8 0

Answer:

2 x^3

Step-by-step explanation:

sqrt(12 x^8)/ sqrt(3x^2)

combine into 1 term

sqrt(12x^8/3x^2)

when dividing exponents with the same base, we subtract

sqrt(12/3  x^(8-2))

sqrt(4 x^6)

sqrt(4)  sqrt(x^6)

2 x^3

boyakko [2]2 years ago
8 0

Answer:

2x^3

Step-by-step explanation:

\frac{\sqrt{12x^8}}{\sqrt{3x^2}}

Both top and bottom have square root . We can take square root in common

\sqrt{\frac{12x^8}{3x^2} }

simplify the exponent using exponential property

a^m / a^n = a^{m-n}, subtract the exponents

\sqrt{\frac{12x^8}{3x^2} }

\sqrt{4x^6}

\sqrt{4} =2

\sqrt{x^6} =\sqrt{x^2 \cdot x^2 \cdot x^2} =x^3

\sqrt{4x^6}=2x^3

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How do you prove each of the following theorems using either a two-column, paragraph, or flow chart proof?
lilavasa [31]

All the theorems are proved as follows.

<h3>What is a Triangle ?</h3>

A triangle is a polygon with three sides , three vertices and three angles.

1. The Triangle sum Theorem

According to the Triangle Sum Theorem, the sum of a triangle's angles equals 180 degrees.

To create a triangle ABC, starting at point A, move 180 degrees away from A to arrive at point B.

We turn 180 degrees from B to C and 180 degrees from C to return to A, giving a total turn of 360 degrees to arrive to A.

180° - ∠A + 180° - ∠B + 180° - ∠C = 360°

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(Hence Proved)

2. Isosceles Triangle Theorem

Considering an isosceles triangle ΔABC

with AB = AC, we have by sine rule;

\rm \dfrac{sinA}{BC} =  \dfrac{sinB}{AC} =  \dfrac{sinC}{AB}\\

as AB = AC

sin B = sin C

angle B = angle C

3.Converse of the Isosceles theorem

Consider an isosceles triangle ΔABC with ∠B= ∠C, we have by sine rule;

\rm \dfrac{sinA}{BC} =  \dfrac{sinB}{AC} =  \dfrac{sinC}{AB}\\

as  ∠B= ∠C ,

AB = AC

4. Midsegment of a triangle theorem

It states that the midsegment of two sides of a triangle is equal to (1/2)of the third side parallel to it.

Given triangle ABC with midsegment at D and F of AB and AC respectively, DF is parallel to BC

In ΔABC and ΔADF

∠A ≅ ∠A

BA = 2 × DA, BC = 2 × FA

Hence;

ΔABC ~ ΔADF (SAS similarity)

BA/DA = BC/FA = DF/AC = 2

Hence AC = 2×DF

5.Concurrency of Medians Theorem

A median of a triangle is a segment whose end points are on vertex of the triangle and the middle point of the side ,the medians of a triangle are concurrent and  the point of intersection is inside the triangle known as Centroid .

Consider a triangle ABC , X,Y and Z are the midpoints of the sides

Since the medians bisect the segment AB into AZ + ZB

BC into BX + XB

AC into AY + YC

Where:

AZ = ZB

BX = XB

AY = YC

AZ/ZB = BX/XB = AY/YC = 1

AZ/ZB × BX/XB × AY/YC = 1 and

the median segments AX, BY, and CZ are concurrent (meet at point within the triangle).

To know more about Triangle

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Step-by-step explanation:

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