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3 years ago

Which expression is equivalent to the expression 3^7x3^-4

Mathematics

1 answer:

kykrilka [37]3 years ago

5 0

Answer:

3^7 * 3^(-4) = 3^(7 - 4) = 3^3 = 27

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Find the value of (-3) Show work Must show work!!

guajiro [1.7K]

Answer:

16/81

Step-by-step explanation:

(-2/3)^4

expand...

-2/3*-2/3*-2/3*-2/3

multiply

numerator: -2*-2*-2*-2=16

denominator: 3*3*3*3=81

Thus the fraction is 16/81

Hope this helps!!!

8 0

2 years ago

25% of x is 30 x=? please help!!!

Nesterboy [21]

25% of 30 can be found by converting 25% to a decimal and multiplying it by 30.

25% = 0.25

30x0.25=7.5

So x is 7.5.

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hope it helps

6 0

3 years ago

The graph of F(x), shown below, has the same shape as the graph of

yarga [219]

Answer:

x^3-1

Step-by-step explanation:

f(x)=g(x)-1 since was shifted down 1 unit x^3-1

f(x)=g(x)+1 would have been a shift up of 1 unit x^3+1

f(x)=g(x-1) shifted right 1 unit (x-1)^3

f(x)=g(x+1) shifted left one unit (x+1)^3

Anyways the answer is f(x)=x^3-1

8 0

3 years ago

Solve for the roots in the equation below. In your final answer, include each of the necessary steps and calculations.

Nostrana [21]

$x^3-27i=0$
$x^3=27i$
$x^3=27e^{i\pi/2}$
$x=\left(27e^{i\pi/2}\right)^{1/3}$
$x=3e^{i(\pi/2+2\pi k)/3}$
$x=3e^{i\pi(4k+1)/6}$

where $k\in\{0,1,2\}$ . This means you have

$x=3e^{i\pi/6}=\dfrac32(\sqrt3+i)$
$x=3e^{i5\pi/6}=\dfrac32(-\sqrt3+i)$
$x=3e^{i9\pi/6}=-3i$

as the solutions to the original equation.

4 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 .

5 0

3 years ago