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antiseptic1488 [7]
3 years ago
7

Find two consecutive whole numbers that the square root of 105 lies between

Mathematics
1 answer:
erastovalidia [21]3 years ago
4 0

Answer: 10 and 11, also -11 and -10

Sqrt 100 is well known to be 10

Sqrt 121 is 11

Step-by-step explanation:

Positive Sqrt is monotone increasing

So Sqrt 105 is between sqrt 100 and sqrt 121


Second answer

Sqrt 100 is less well known to also be -10

Negative sqrt is monotone decreasing

Sqrt 121 is also -11

Sqrt 105 is also between -121 and -100

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5. Find the first, fourth, and tenth terms of the arithmetic
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The first, fourth and tenth terms of the arithmetic sequence is -6, -\frac{27}{5} and -\frac{21}{5}

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The given rule for the arithmetic sequence is A(n)=-6+(n-1)(\frac{1}{5} )

We need to determine the first, fourth and tenth terms of the sequence.

To find the first, fourth and tenth terms, let us substitute n=1,4,10 in the general rule for the arithmetic sequence.

To find the first term, substitute n=1 in A(n)=-6+(n-1)(\frac{1}{5} ) , we get,

A(1)=-6+(1-1)(\frac{1}{5} )

A(1)=-6+(0)(\frac{1}{5} )

A(1)=-6

Thus, the first term of the arithmetic sequence is -6.

To find the fourth term, substitute n=4 in A(n)=-6+(n-1)(\frac{1}{5} ) , we get,

A(2)=-6+(4-1)(\frac{1}{5} )

A(2)=-6+(3)(\frac{1}{5} )

A(2)=\frac{-30+3}{5}

A(2)=\frac{-27}{5}

Thus, the fourth term of the arithmetic sequence is -\frac{27}{5}

To find the tenth term, substitute n=10 in A(n)=-6+(n-1)(\frac{1}{5} ) , we get,

A(10)=-6+(10-1)(\frac{1}{5} )

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A(10)=-6+\frac{9}{5}

A(10)=-\frac{21}{5}

Thus, the tenth term of the arithmetic sequence is -\frac{21}{5}

Hence, the first, fourth and tenth terms of the arithmetic sequence is -6, -\frac{27}{5} and -\frac{21}{5}

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