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3 years ago

10

How many solutions are there to a cubed root problem

1 answer:

Gelneren [198K]3 years ago

6 0

Answer:

Cubic equations, which are polynomial equations of the third degree (meaning the highest power of the unknown is 3) can always be solved for their three solutions in terms of cube roots and square roots (although simpler expressions only in terms of square roots exist for all three solutions

Step-by-step explanation:

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Which number is a rational number? A. B. C. D.

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A card is drawn from a well-shuffled deck of 52 cards. find the probability that the card is a red card or a 10.

harina [27]

Total number of cards = 52

-

<u>Find total number of red card or a 10:</u>

Total number of cards that are red = 26

Total number of cards that are 10 but not red = 2

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P(red card or a 10) = 28/52 = 7/13

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3 years ago

Explain me PLEASEEEE!!!

kirza4 [7]

Answer:

$\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10}}=\frac{a^{3}}{a^{10}}=a^{-7}=\frac{1}{a^{7}}$

Step-by-step explanation:

Let us revise the properties of exponents

$a^{m}.a^{n}=a^{m+n}$

$\frac{a^{m}}{a^{n}}=a^{m-n}$

$(a^{m})^{n}=a^{m.n}$

$a^{-m}=\frac{1}{a^{m} }$

Let us use these properties to solve the question

→ By using the 3rd property above

∵ $(a^{2})^{3}=a^{2.3}=a^{6}$

∴ $\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10} }$

→ By using the 1st property above

∵ $a^{6}.a^{-3}=a^{6+-3}=a^{6-3}=a^{3}$

∴ $\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10}}=\frac{a^{3}}{a^{10}}$

→ By using the 2nd property above

∵ $\frac{a^{3}}{a^{10}}=a^{3-10}=a^{-7}$

∴ $\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10}}=\frac{a^{3}}{a^{10}}=a^{-7}$

→ By using the 4th property above

∵ $a^{-7}=\frac{1}{a^{7}}$

∴ $\frac{(a^{2})^{3}.a^{-3}}{a^{10}}=\frac{a^{6}.a^{-3}}{a^{10}}=\frac{a^{3}}{a^{10}}=a^{-7}=\frac{1}{a^{7}}$

8 0

3 years ago