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3 years ago

Which expression is equivalent to (4x^(3)y^(5))(3x^(5)y)^(2)

Mathematics

1 answer:

marshall27 [118]3 years ago

8 0

Answer:

$(4x^3y^5)(3x^5y)^2 = 36*x^{13}y^7$

Step-by-step explanation:

Given

$(4x^3y^5)(3x^5y)^2$

Required

The equivalent expression

We have:

$(4x^3y^5)(3x^5y)^2$

Expand

$(4x^3y^5)(3x^5y)^2 = 4x^3y^5*9x^{10}y^2$

Further expand

$(4x^3y^5)(3x^5y)^2 = 4*9*x^3*x^{10}y^5*y^2$

Apply laws of indices

$(4x^3y^5)(3x^5y)^2 = 36*x^{13}y^7$

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A flying disc has a circumference of 75.36 centimeters. What is the area of the flying disc? (Use 3.14 for .)

Hatshy [7]

The area of flying disk is 452.16 square centimeters

Step-by-step explanation:

The disc is in circular shape so we will use the formulas for circle in this question.

Given

Circumference = C = 75.36 centimeters

We have to find the area of the disc for which we have to find the radius first.

Let r be the radius

Then

Circumference is given by:

$C=2\pi r$

putting the values

$75.36 = 2*3.14*r\\75.36 = 6.28r$

Dividing both sides by 6.28

$\frac{6.28r}{6.28} = \frac{75.36}{6.28}\\r = 12\ cm$

The area of circle is given by:

$A = \pi r^2$

Putting the values

$A = 3.14 * (12)^2\\A = 3.14*144\\A =452.16\ cm^2$

Hence,

The area of flying disk is 452.16 square centimeters

Keywords: Circle, area

Learn more about circle at:

#LearnwithBrainly

5 0

3 years ago

Prove that H c G is a normal subgroup if and only if every left coset is a right coset, i.e., aH = Ha for all a e G

Kaylis [27]

$\Rightarrow$

Suppose first that $H\subset G$ is a normal subgroup. Then by definition we must have for all $a\in H$ , $xax^{-1} \in H$ for every $x\in G$ . Let $a\in G$ and choose $(ab)\in aH$ ( $b\in H$ ). By hypothesis we have $aba^{-1} =abbb^{-1}a^{-1}=(ab)b(ab)^{-1} \in H$ , i.e. $aba^{-1}=c$ for some $c\in H$ , thus $ab=ca \in Ha$ . So we have $aH\subset Ha$ . You can prove $Ha\subset aH$ in the same way.

$\Leftarrow$

Suppose $aH=Ha$ for all $a\in G$ . Let $h\in H$ , we have to prove $aha^{-1} \in H$ for every $a\in G$ . So, let $a\in G$ . We have that $ha^{-1} =a^{-1}h'$ for some $h'\in H$ (by the hypothesis). hence we have $aha^{-1}=h' \in H$ . Because $a$ was chosen arbitrarily we have the desired .