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2 years ago

11

a store sells jump ropes. each jump rope costs the same amount. during a sale, the store reduces the price of each jump rope by

$1.25. jada spends $15.08 on 2 jump ropes at the sale price.

1 answer:

fiasKO [112]2 years ago

4 0

Answer:

she spent $7.45 on each rope at the sale price. The original price would have been $8.70

Step-by-step explanation:

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3x + 1 − 4x3 + 6x6 −2x2 standard form

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Answer:

$(x^{-3} )^{2}$

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Step-by-step explanation:

$(x^{3} x^{-6} )^{2}$ is the expression given to be solved.

First of all let us have a look at <u>3 formulas</u>:

$1.\ p^a \times p^b = p^{(a+b)}\\2.\ (p^a \times q^b)^c = (p^{a})^c \times (q^{b})^c\\3.\ (p^a)^b = p^{a\times b}$

Both the formula can be applied to the expression( $(x^{3} x^{-6} )^{2}$ ) during the first step while solving it.

<u>Applying formula (1):</u>

$(x^{3} x^{-6} )^{2}$

Comparing the terms of $(x^{3} x^{-6} )$ with $p^a \times p^b$

$p=x, a =3, b=-6$

$\Rightarrow x^{3+(-6)}\\\Rightarrow x^{3-6}\\\Rightarrow x^{-3}$

So, $(x^{3} x^{-6} )^{2}$ is reduced to $(x^{-3} )^{2}$

<u>Applying formula (2):</u>

Comparing the terms of $(x^{3} x^{-6} )^{2}$ with $(p^a \times q^b)^c$

$p=q=x, a =3, b=-6, c=2$

$\Rightarrow (x^{3})^2\times (x^{-6})^2\\\text{Applying Formula (3)}\\x^6 x^{-12}$

So, $(x^{3} x^{-6} )^{2}$ is reduced to $x^6 x^{-12}$ .

So, the answers can be:

$(x^{-3} )^{2}$

$x^6 x^{-12}$

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