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1 year ago

Cube roots till 30... pleaseeee!

Mathematics

2 answers:

masha68 [24]1 year ago

5 0

Answer:

1, 8, 27?

Wasn't sure if you were asking for the cube numbers instead?

BabaBlast [244]1 year ago

4 0

<em>Answer:</em>

1³ = 1
2³ = 8
3³ = 27
4³ = 64
5³ = 125
6³ = 216
7³ = 343
8³ = 512
9³ = 729
10³ = 1000
11³ = 1331
12³ = 1728
13³ = 2197
14³ = 2744
15³ = 3375
16³ = 4096
17³ = 4913
18³ = 5832
19³ = 6859
20³ = 8000
21³ = 9261
22³ = 10648
23³ = 12167
24³ = 13824
25³ = 15625
26³ = 17576
27³ = 19683
28³ = 21952
29³ = 24389
30³ = 27000

If this isn't what you wanted, the cube roots are the first three.
Hope this helps !!

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Use properties of limits and algebraic methods to find the limit, if it exists. (If the limit is infinite, enter '[infinity]' or

soldi70 [24.7K]

Answer:

The limit of this function does not exist.

Step-by-step explanation:

$\lim_{x \to 3} f(x)$

$f(x)=\left \{ {{9-3x} \quad if \>{x \>< \>3} \atop {x^{2}-x }\quad if \>{x\ \geq \>3 }} \right.$

To find the limit of this function you always need to evaluate the one-sided limits. In mathematical language the limit exists if

$\lim_{x \to a^{-}} f(x) = \lim_{x \to a^{+}} f(x) =L$

and the limit does not exist if

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Evaluate the one-sided limits.

The left-hand limit

$\lim_{x \to 3^{-} } 9-3x= \lim_{x \to 3^{-} } 9-3*3=0$

The right-hand limit

$\lim_{x \to 3^{+} } x^{2} -x= \lim_{x \to 3^{+} } 3^{2}-3 =6$

Because the limits are not the same the limit does not exist.

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3 years ago

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sertanlavr [38]

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2 years ago

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Draw a model of square root of 12 using perfect squares

Shkiper50 [21]

Answer:

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Step-by-step explanation:

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$= \sqrt{2^2\times 3}\\\\= 2\sqrt{3}\\\\$

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3 years ago

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Cube roots till 30... pleaseeee!​

Cube roots till 30... pleaseeee!