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2 years ago

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The power 3 Superscript negative 3 equals StartFraction 1 Over 27 EndFraction . Which expression is equivalent to 3 Superscript

negative 3?
StartFraction 1 Over 3 squared EndFraction
StartFraction 1 Over 3 cubed EndFraction
StartFraction 1 Over 9 cubed EndFraction
StartFraction 1 Over 3 Superscript 9 EndFraction

2 answers:

LiRa [457]2 years ago

8 0

Answer:

B. $\frac{1}{3^3}$

Step-by-step explanation:

We have been given an expression $3^{-3}=\frac{1}{27}$ . We are asked to find an equivalent expression to $3^{-3}$ .

We will use exponent properties to solve our given problem.

We will use exponent property $a^{-b}=\frac{1}{a^b}$ to rewrite our given expression.

Upon comparing our expression with exponent property, we can see that $a=3$ and $-b=-3$ .

$3^{-3}=\frac{1}{3^3}$

Therefore, expression $\frac{1}{3^3}$ in equivalent to $3^{-3}$ and option B is the correct choice.

SpyIntel [72]2 years ago

4 0

Answer:

Option a: \frac{m^{5} }{162n} is the equivalent expression.

Explanation:

The expression is \frac{(3m^{-2} n)^{-3}}{6mn^{-2} } where m\neq 0, n\neq 0

Let us simplify the expression, to determine which expression is equivalent from the four options.

Multiplying the powers, we get,

\frac{3^{-3}m^{6} n^{-3}}{6mn^{-2} }

Cancelling the like terms, we have,

\frac{3^{-3}m^{5} n^{-1}}{6 }

This equation can also be written as,

\frac{m^{5}}{3^{3}6 n^{1} }

Multiplying the terms in denominator, we have,

\frac{m^{5} }{162n}

Thus, the expression which is equivalent to \frac{(3m^{-2} n)^{-3}}{6mn^{-2} } is \frac{m^{5} }{162n}

Hence, Option a is the correct answer. 23

Step-by-step explanation:

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Answer

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Step-by-step explanation:

Before we proceed let us re write correctly the height equation which in correct form reads:

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Thus Eqn. (2) becomes:

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<u>Josh:</u>

We know that $v_{o}=48ft/s$

Thus Eqn. (2) becomes:

$h(t)=-16t^2+48t+40$ Eqn. (4).

<em><u>Now since we want to find whose textbook reaches the ground first and by how many seconds we need to solve each equation (i.e. Eqns. (3) and (4)) at </u></em> $h(t)=0$ <em><u>. Now since both are quadratic equations we will solve one showing the full method which can be repeated for the other one. </u></em>

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$t_{1,2}=\frac{-b+/-\sqrt{b^2-4ac} }{2a} \\t_{1,2}=\frac{-60+/-\sqrt{60^2-4(-16)(40)} }{2(-16)} \\t_{1,2}=\frac{-60+/-\sqrt{6160} }{-32} \\t_{1,2}=\frac{15+/-\sqrt{385} }{8}\\\\t_{1}=4.3276 sec\\t_{2}=-0.5776 sec$

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$t_{1}=3.6794sec\\t_{2}=-0.6794sec$

Thus again we reject the negative and keep the positive solution, so Josh's book took $t=3.6794$ seconds to reach the ground.

Then we can find the difference between Ben and Josh times as

$t_{Ben}-t_{Josh}= 4.3276 - 3.6794 = 0.6482$

So to answer the original question:

<em>Whose textbook reaches the ground first and by how many seconds?</em>

Josh's textbook reached the ground first
Josh's textbook reached the ground first by a difference of $t=0.6482$

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